Proof of Basic Properties of the Exponential Function

Notation. For $u\in\mathbb{R}$ and $k\in\mathbb{N}$ let $t_{k}(u)=u^{k-1}/(k-1)!$, with powers, factorials and $k-1$ as in The Exponential Series Converges Absolutely, and Its Sum Is a Map on ℝ; for each $u\in\mathbb{R}$, $(t_{k}(u))_{k\in\mathbb{N}}$ is the sequence of The Exponential Series Converges Absolutely, and Its Sum Is a Map on ℝ (i). By The Exponential Series Converges Absolutely, and Its Sum Is a Map on ℝ (ii), the series $\sum_{k=1}^{\infty}t_{k}(u)$ converges absolutely, and in particular converges, and by The Real Exponential Function (i) and The Real Exponential Function (ii) its sum is $\exp(u)$; so its partial sums $\sum_{k=1}^{N}t_{k}(u)$, a sequence by Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (i), converge to $\exp(u)$, by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence (ii). Natural numbers with zero are read in $\mathbb{R}$ by The Real Line: Standing Notation and Background for Calculus on Intervals (vii); their equality, sums, differences, products and order agree with those formed in $\mathbb{R}$ by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vi) (for the order, see also Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (i)), and the identities and inequalities between natural numbers below are checked in $\mathbb{R}$ in this way. Every $m\in\mathbb{N}$ satisfies $1\le m$ by Arithmetic and Order of the Natural Numbers (x), also in $\mathbb{R}$ by Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (ii); an $m\in\mathbb{N}_{0}$ with $1\le m$ lies in $\mathbb{N}$ by Elements of a Segment Are Natural Numbers, and Mixed Transitivity of the Order on the Natural Numbers (i); and for $m\in\mathbb{N}_{0}$, in $\mathbb{R}$, $1\le m!\le(m+1)!$ and $m!>0$ by Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (iii).

The first two terms. For $u\in\mathbb{R}$: $t_{1}(u)=u^{0}/0!=1$ by Powers with Exponents in the Natural Numbers with Zero (ii) and The Factorial: Recursion, Positivity and Bounds by Powers (i); and as $2=1+1$ by Arithmetic and Order of the Natural Numbers (xiv), $2-1=1$, $u^{1}=u$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii), and $1!=1\cdot0!=1$ by The Factorial: Recursion, Positivity and Bounds by Powers (i), so $t_{2}(u)=u$. As $[1]=\{1\}$, Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (i) and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (vii) give

$$\sum_{k=1}^{2}t_{k}(u)=t_{1}(u)+t_{2}(u)=1+u.\tag{1}$$

Proof of (i). Let $k\in\mathbb{N}$ with $k\neq1$. Then $k-1\neq0$, since $k=1+(k-1)$; so $k-1\in\mathbb{N}$, $1\le k-1$, and $0^{k-1}=0$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii), whence $t_{k}(0)=0$. As $t_{1}(0)=1$ and $1\in[N]$, Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality (iv) gives $\sum_{k=1}^{N}t_{k}(0)=1$ for every $N\in\mathbb{N}$. The constant sequence $(1)_{N\in\mathbb{N}}$ (Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (v)) converges to $1$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (i); it is the sequence of partial sums of $\sum_{k=1}^{\infty}t_{k}(0)$, which converges to $\exp(0)$ (see the Notation paragraph), so $\exp(0)=1$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (ii), as in The Limit of a Convergent Sequence (i).

Proof of (ii). Let $u,v\in\mathbb{R}$. Apply The Cauchy Product of Two Absolutely Convergent Series with $a_{k}=t_{k}(u)$ and $b_{k}=t_{k}(v)$, whose series converge absolutely by The Exponential Series Converges Absolutely, and Its Sum Is a Map on ℝ (ii), and let $D_{n}$ and $c_{n}$ be as there. Let $n\in\mathbb{N}$ and $m=n-1\in\mathbb{N}_{0}$. We show $c_{n}=(u+v)^{m}/m!=t_{n}(u+v)$.

For $k\in\{0,\dots,m\}$ the difference $m-k\in\mathbb{N}_{0}$ is defined, and in $\mathbb{R}$ we have $1\le k+1\le m+1=n$, $1\le(m-k)+1\le n$ and $(k+1)+((m-k)+1)=m+2=n+1$. Hence $\varphi(k)=(k+1,(m-k)+1)$ lies in $D_{n}$, and $\varphi:\{0,\dots,m\}\to D_{n}$ is a map. It is injective, since $k+1=k'+1$ gives $k=k'$. It is surjective: for $(i,j)\in D_{n}$ let $k=i-1\in\mathbb{N}_{0}$; in $\mathbb{R}$, $k+(j-1)=(i+j)-2=n-1=m$ with $j-1\ge0$, so $k\le m$, $m-k=j-1$ and $\varphi(k)=(i,j)$. So $\varphi$ is a bijection. Since $(k+1)-1=k$ and $((m-k)+1)-1=m-k$, Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (iii), Sums and Products over a Finite Set and over an Interval (iv) and Finite Sums of Reals: Congruence, the Product Map (x,y) ↦ f(x)h(y), and the Sequence of Diagonal Sums (i), for the reindexed terms, which agree on $\{0,\dots,m\}$, give

$$c_{n}=\sum_{(i,j)\in D_{n}}t_{i}(u)\,t_{j}(v)=\sum_{k=0}^{m}t_{k+1}(u)\,t_{(m-k)+1}(v)=\sum_{k=0}^{m}\frac{u^{k}v^{m-k}}{k!\,(m-k)!}.$$

By the binomial theorem The Binomial Theorem in a Commutative Ring (i) in the commutative ring $\mathbb{R}$, with the binomial coefficients $\binom{m}{k}$, $(u+v)^{m}=\sum_{k=0}^{m}\binom{m}{k}u^{k}v^{m-k}$. For $k\le m$, $k!\,(m-k)!\,\binom{m}{k}=m!$ by Binomial Coefficients: Existence and Uniqueness by Pascal's Recursion, Vanishing above the Diagonal, Diagonal and First Values, the Factorial Formula and Symmetry (v) and The Binomial Coefficient (i), an identity in $\mathbb{N}_{0}$ that holds in $\mathbb{R}$, the image in $\mathbb{R}$ of a product being the product of the images by The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products (iii); as the factorials are nonzero (Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (iii)), $\binom{m}{k}/m!=1/(k!\,(m-k)!)$. Hence, by Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality (i),

$$\frac{(u+v)^{m}}{m!}=\sum_{k=0}^{m}\frac{\binom{m}{k}}{m!}\,u^{k}v^{m-k}=\sum_{k=0}^{m}\frac{u^{k}v^{m-k}}{k!\,(m-k)!}=c_{n}.$$

So $c_{n}=t_{n}(u+v)$ for every $n\in\mathbb{N}$, and the sequences $(c_{n})$ and $(t_{n}(u+v))$ are equal by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv). By The Cauchy Product of Two Absolutely Convergent Series (ii), their series converges with sum $\exp(u)\exp(v)$, so its partial sums converge to $\exp(u)\exp(v)$ by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence (ii); they also converge to $\exp(u+v)$ (see the Notation paragraph), so $\exp(u+v)=\exp(u)\exp(v)$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (ii).

Reciprocals. Let $u\in\mathbb{R}$. By (ii) and (i), $\exp(u)\exp(-u)=\exp(u+(-u))=\exp(0)=1$. So $\exp(u)\neq0$, as $0\cdot\exp(-u)=0\neq1$ (Rules of Arithmetic and Order in an Ordered Field (i), Rules of Arithmetic and Order in an Ordered Field (vii)), and multiplying by the reciprocal $\exp(u)^{-1}$ (Negatives, Reciprocals and the Strict Order in an Ordered Field, as Values (ii)) gives

$$\exp(-u)=\frac{1}{\exp(u)}\qquad(u\in\mathbb{R}).\tag{2}$$

This is the identity of (iv).

Proof of (v) for $u\ge0$, and of (iii). Let $u\ge0$. For $k\in\mathbb{N}$, $u^{k-1}\ge0$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (vi) and $(k-1)!>0$, so $1/(k-1)!>0$ by Rules of Arithmetic and Order in an Ordered Field (viii), and $t_{k}(u)=u^{k-1}\cdot(1/(k-1)!)\ge0\cdot(1/(k-1)!)=0$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (iv) and Rules of Arithmetic and Order in an Ordered Field (i). By Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (i) and (1), $1+u=\sum_{k=1}^{2}t_{k}(u)\le\exp(u)$. In particular $\exp(u)\ge1+u\ge1>0$ for every $u\ge0$. If $u<0$, then $-u>0$ (Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (iii)), so $\exp(-u)>0$ and, by (2) applied to $-u$, $\exp(u)=1/\exp(-u)>0$ by Rules of Arithmetic and Order in an Ordered Field (viii). As $u<0$ or $0\le u$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vii), (iii) follows.

Proof of (v) for $u<0$. If $u\le-1$, then $1+u\le0<\exp(u)$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (ii) and (iii). Let $-1<u<0$ and $w=-u$, so $0<w<1$ and $|w|<1$. For $k\in\mathbb{N}$, $w^{k-1}>0$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (vi), as $w>0$. Also $0<1/(k-1)!\le1$: if $(k-1)!=1$ this is $0<1$ (Rules of Arithmetic and Order in an Ordered Field (vii)), and otherwise $1<(k-1)!$, as $1\le(k-1)!$, by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (v), so $0<1/(k-1)!<1/1=1$ by Rules of Arithmetic and Order in an Ordered Field (viii). Multiplying $0\le1/(k-1)!\le1$ by $w^{k-1}>0$ gives $0\le t_{k}(w)\le w^{k-1}$ by Rules of Arithmetic and Order in an Ordered Field (vi) and Rules of Arithmetic and Order in an Ordered Field (i). The sequence $(w^{k-1})_{k\in\mathbb{N}}$ is that of Power Series: the Sequence of Terms, the Coefficients 1/(k-1)!, and the Sum as a Map on the Set of Convergence (i) for $u=w$ and the constant coefficient sequence $1$ of Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (v), as $1\cdot w^{k-1}=w^{k-1}$. By Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (v), $\sum_{k=1}^{\infty}w^{k-1}$ converges with sum $1/(1-w)$, so Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (ii) gives $\exp(w)\le1/(1-w)$. Multiplying by $(1-w)/\exp(w)>0$ (Rules of Arithmetic and Order in an Ordered Field (vi), using (iii) and $1-w>0$) gives $1-w\le1/\exp(w)$. By (2) applied to $w$, $\exp(u)=\exp(-w)=1/\exp(w)\ge1-w=1+u$.

Proof of (vi). Let $h\in\mathbb{R}$ with $|h|\le1$. By Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping (iv) with $p=2$, applied to the shifted sequence $(t_{2+k}(h))_{k\in\mathbb{N}}$ of Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (ii), and (1),

$$\exp(h)=\sum_{k=1}^{2}t_{k}(h)+\sum_{k=1}^{\infty}t_{2+k}(h)=1+h+\sum_{k=1}^{\infty}t_{2+k}(h),$$

the last series converging. Let $k\in\mathbb{N}$. Then $(2+k)-1=k+1$, so $t_{2+k}(h)=h^{k+1}/(k+1)!$. By Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (vi) and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (i) (with $(k-1)+2=k+1$), $|h^{k+1}|=|h|^{k+1}=|h|^{k-1}|h|^{2}$; here $|h|^{k-1}\le1$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ix), as $0\le|h|\le1$ (Differences, Quotients and Absolute Values of Real Numbers, Read in the Type Real (iii)), and $|h|^{2}=|h^{2}|=h^{2}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (vi), as $h^{2}=h\cdot h\ge0$ by Rules of Arithmetic and Order in an Ordered Field (vii). By Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (iii) applied to $k-1$ and to $k$, $1\le(k-1)!\le k!\le(k+1)!$ in $\mathbb{R}$, so $1/(k+1)!\le1/(k-1)!$: the two are equal if $(k-1)!=(k+1)!$, and otherwise $(k-1)!<(k+1)!$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (v) and Rules of Arithmetic and Order in an Ordered Field (viii) applies. Hence $|h|^{k+1}=|h|^{k-1}h^{2}\le1\cdot h^{2}=h^{2}$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (iv), as $0\le h^{2}$; and, as $0\le|h|^{k+1}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (vi) and $0<1/(k+1)!$ by Rules of Arithmetic and Order in an Ordered Field (viii), multiplying $0\le|h|^{k+1}\le h^{2}$ and $0\le1/(k+1)!\le1/(k-1)!$ gives $|h|^{k+1}/(k+1)!\le h^{2}/(k-1)!$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (vii). With $1^{k-1}=1$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii)) and Rules of Arithmetic and Order in an Ordered Field (x),

$$|t_{2+k}(h)|=\frac{|h|^{k+1}}{(k+1)!}\le\frac{h^{2}}{(k-1)!}=h^{2}\,t_{k}(1).$$

By Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping (i) and The Real Exponential Function (iii), $\sum_{k=1}^{\infty}h^{2}t_{k}(1)$, the series of the sequence $(h^{2}t_{k}(1))_{k\in\mathbb{N}}$ of Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (v), converges with sum $h^{2}\exp(1)=e\,h^{2}$. Hence, by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (iv),

$$|\exp(h)-1-h|=\Big|\sum_{k=1}^{\infty}t_{2+k}(h)\Big|\le e\,h^{2}.$$

Proof of (vii). Let $I$ be an interval containing two distinct points, $f=\exp|_{I}$, so $f(s)=\exp(s)$ for $s\in I$, and let $x\in I$. We show that $\exp(x)$ is a derivative of $f$ at $x$. Let a real $\varepsilon>0$ be given. Put $K=e\exp(x)+1$; as $e=\exp(1)>0$ and $\exp(x)>0$ by (iii), $e\exp(x)>0\cdot\exp(x)=0$ by Rules of Arithmetic and Order in an Ordered Field (vi) and Rules of Arithmetic and Order in an Ordered Field (i), so $K>1>0$. Now choose $\delta=\min\{1,\varepsilon/K\}$, which depends on $\varepsilon$ and $x$ only; $\varepsilon/K>0$ by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (ii) and Rules of Arithmetic and Order in an Ordered Field (viii), so $\delta>0$, $\delta\le1$ and $\delta\le\varepsilon/K$ by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (viii). Let $t\in I$ with $0<|t-x|<\delta$, and put $h=t-x$ (Differences, Quotients and Absolute Values of Real Numbers, Read in the Type Real (i)); then $h\neq0$ and $|h|<\delta\le1$. By (ii), $f(t)=\exp(x+h)=\exp(x)\exp(h)$, so

$$\frac{f(t)-f(x)}{t-x}-\exp(x)=\exp(x)\,\frac{\exp(h)-1-h}{h}.$$

By Rules of Arithmetic and Order in an Ordered Field (x), (vi) and $h^{2}=|h|^{2}$, its absolute value is at most $\exp(x)\,e\,h^{2}/|h|=e\exp(x)\,|h|\le K|h|<K\delta\le K\cdot\varepsilon/K=\varepsilon$ (Rules of Arithmetic and Order in an Ordered Field (vi), Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (iv)). So $\exp(x)$ is a derivative of $f$ at $x$; hence $f$ is differentiable at $x$ with $f'(x)=\exp(x)$, the derivative being unique as stated there. As $x\in I$ was arbitrary, $f$ is differentiable by Derivative of a Real Function on an Interval (iii).

Proof of (viii). $\mathbb{R}$ is an interval containing the distinct points $0$ and $1$, and $\exp|_{\mathbb{R}}=\exp$ by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iii). By (vii) with $I=\mathbb{R}$, $\exp$ is differentiable at every $x\in\mathbb{R}$, hence continuous at $x$ by The Increment Form of the Derivative, and Differentiability Implies Continuity (ii). So $\exp$ is continuous on $\mathbb{R}$, by The Real Line: Standing Notation and Background for Calculus on Intervals (vi).

Proof of (ix). Let $u<v$ and $w=v-u$, so $w>0$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (i). By (v), $\exp(w)\ge1+w>1$. By (ii), $\exp(v)=\exp(u+w)=\exp(u)\exp(w)$, and as $\exp(u)>0$ by (iii), $\exp(u)=\exp(u)\cdot1<\exp(u)\exp(w)=\exp(v)$ by Rules of Arithmetic and Order in an Ordered Field (vi).

Proof of (x). Let a real $\varepsilon>0$ be given, and choose $M=1/\varepsilon$, so $M>0$ by Rules of Arithmetic and Order in an Ordered Field (viii). Let $u<-M$ and $w=-u$; then $w>M>0$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (iii). By (v), $\exp(w)\ge1+w>w>M=1/\varepsilon>0$. By (2) applied to $w$, Rules of Arithmetic and Order in an Ordered Field (viii) and Rules of Arithmetic and Order in an Ordered Field (iii), $\exp(u)=1/\exp(w)<1/(1/\varepsilon)=\varepsilon$.

Proof of (xi). Let $g:\mathbb{R}\to(0,\infty)$, $g(u)=\exp(u)$. Injective: let $u,u'\in\mathbb{R}$ with $u\neq u'$. By Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vi), $u<u'$ or $u'<u$, so $\exp(u)<\exp(u')$ or $\exp(u')<\exp(u)$ by (ix), and in either case $g(u)\neq g(u')$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (ii). Surjective: let $s\in(0,\infty)$, so $s>0$. First let $s\ge1$; then $0<s$. The restriction of $\exp$ to the closed interval $[0,s]$ is continuous on $[0,s]$, by (viii) and The Real Line: Standing Notation and Background for Calculus on Intervals (vi), and $\exp(0)=1\le s<1+s\le\exp(s)$ by (i) and (v). So Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals (i) gives $c\in[0,s]$ with $\exp(c)=s$. Now let $0<s<1$. Then $1/s>1$ by Rules of Arithmetic and Order in an Ordered Field (viii), so the first case gives $c\in\mathbb{R}$ with $\exp(c)=1/s$, and by (2) and Rules of Arithmetic and Order in an Ordered Field (iii), $\exp(-c)=1/(1/s)=s$. As $s<1$ or $1\le s$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vii), $g$ is surjective, and so it is bijective.