Proof of Basic Properties of the Exponential Function

Claim 1. $\exp(0)=1$ directly from the series of The Real Exponential Function with the convention $0^{0}=1$. For multiplicativity, form the Cauchy product of the two absolutely convergent series: the family $u^{j}v^{k}/(j!k!)$ over pairs $(j,k)$ is absolutely summable (its absolute sums are bounded by the product of the two absolute-value series), so it may be summed by diagonals — for series of nonnegative terms every rearrangement has the same supremum of finite partial sums, and the general case follows by splitting into positive and negative parts. Grouping by $j+k=l$ and using the binomial expansion $(u+v)^{l}=\sum_{j\le l}\binom{l}{j}u^{j}v^{l-j}$ with $\binom{l}{j}=l!/(j!(l-j)!)$ (Factorial of a Natural Number) gives $\exp(u)\exp(v)=\sum_{l}(u+v)^{l}/l!=\exp(u+v)$.

Claim 2. By claim 1, $\exp(u)\exp(-u)=\exp(0)=1$, so $\exp(u)\ne 0$ and $\exp(-u)=1/\exp(u)$. For $u\ge 0$ the series has nonnegative terms and constant term $1$, so $\exp(u)\ge 1>0$; for $u<0$, $\exp(u)=1/\exp(-u)>0$.

Claim 3. From the series, for $|s|\le 1$,

so $\bigl(\exp(s)-1\bigr)/s\to 1$ as $s\to 0$. By claim 1, $\bigl(\exp(u+s)-\exp(u)\bigr)/s=\exp(u)\bigl(\exp(s)-1\bigr)/s\to\exp(u)$, so $\exp$ has derivative $\exp'=\exp$ at every point. Iterating, all higher derivatives exist and equal $\exp$, which is continuous (differentiability at a point implies continuity there, directly from the definition of the derivative and limit arithmetic); hence $\exp$ is a smooth map on $\mathbb{R}=\mathbb{R}^1$.

Claim 4. $\exp(u)\ge 1+u$ for $u\ge 0$: all omitted series terms are nonnegative. Strict monotonicity: for $u<v$, claims 1 and 2 give $\exp(v)-\exp(u)=\exp(u)\bigl(\exp(v-u)-1\bigr)\ge\exp(u)(v-u)>0$. Decay at $-\infty$: given $\varepsilon>0$ choose $M>0$ with $1/(1+M)<\varepsilon$; for $u<-M$, $\exp(-u)\ge 1+(-u)>1+M$, so $\exp(u)=1/\exp(-u)<\varepsilon$.

Claim 5. Injectivity is strict monotonicity (claim 4). For surjectivity onto $(0,\infty)$, let $t\ge 1$ (the case $0<t<1$ follows by applying the result to $1/t$ and using claim 2). The set $S=\{u\in[0,t]:\exp(u)\le t\}$ contains $0$ and is bounded above by $t$, so it has a least upper bound $s$ by Least Upper Bound Property of the Real Numbers. Since $\exp$ is continuous at $s$ (claim 3): if $\exp(s)<t$, then $s<t$ (as $\exp(t)\ge 1+t>t$) and $\exp<t$ on a neighborhood of $s$, contradicting that $s$ is an upper bound of $S$; if $\exp(s)>t$, then $\exp>t$ on a neighborhood of $s$, contradicting that $s$ is the least upper bound (points of $S$ would be excluded from that neighborhood, giving a smaller upper bound). Hence $\exp(s)=t$, and $\exp$ is a bijection from $\mathbb{R}$ onto $(0,\infty)$ (the image is contained in $(0,\infty)$ by claim 2). $\blacksquare$