Proof of Properties of Natural Number Powers in a Field

By Natural Number Power of an Element of a Field, $c^{n}$ is the finite product $\prod_{k=1}^{n}a_{k}$ of the map $a$ on the initial segment $[n]$ whose value at every index is $c$; write $b$ and $e$ for the corresponding constant maps with values $d$ and $c\,d$. In claims 1 to 5 below, references to numbered claims are to Properties of Finite Products.

Claim 1. Both identities are claim 1, since the constant map on $[S(n)]$ with value $c$ restricts to the constant map on $[n]$ with value $c$, so that $\prod_{k=1}^{n}$ may be read for either family.

Claim 2. We argue by induction on $n$, using the induction principle for the natural numbers. By claim 1, $1^{1}=1$. If $1^{n}=1$, then $1^{S(n)}=1^{n}\cdot1=1\cdot1=1$ by claim 1 and the defining property of the multiplicative identity.

Claim 3. For every $k\in[n]$ we have $e_{k}=c\,d=a_{k}\,b_{k}$, so claim 2 gives

$$(c\,d)^{n}=\prod_{k=1}^{n}\bigl(a_{k}b_{k}\bigr)=\Bigl(\prod_{k=1}^{n}a_{k}\Bigr)\Bigl(\prod_{k=1}^{n}b_{k}\Bigr)=c^{n}\,d^{n}.$$

Claim 4. By claim 4, $\prod_{k=1}^{n}a_{k}=0$ holds if and only if $a_{i}=0$ for some $i\in[n]$. Every value of $a$ is $c$, and $[n]$ is nonempty since $1\in[n]$, so this condition holds if and only if $c=0$.

Claim 5. Since $0\le a_{k}$ for every $k\in[n]$, the first part of claim 5 gives $0\le\prod_{k=1}^{n}a_{k}=c^{n}$. If in addition $c\le d$, then $a_{k}\le b_{k}$ for every $k\in[n]$, so the second part of claim 5 gives $c^{n}=\prod_{k=1}^{n}a_{k}\le\prod_{k=1}^{n}b_{k}=d^{n}$.

The remaining two claims cite clauses of the present statement as "clause $k$", to keep them apart from the claims of the finite-product lemma.

Claim 6. Fix $m\in\mathbb{N}$ and let $E$ be the set of $j\in\mathbb{N}$ with $c^{m+j}=c^{m}c^{j}$. By Natural Numbers, $m+1=S(m)$ and $m+S(j)=S(m+j)$ for every $j$. Clause 1 gives $c^{m+1}=c^{S(m)}=c^{m}c=c^{m}c^{1}$, so $1\in E$. If $j\in E$, then clause 1 and associativity of multiplication give

$$c^{m+S(j)}=c^{S(m+j)}=c^{m+j}\,c=\bigl(c^{m}c^{j}\bigr)c=c^{m}\bigl(c^{j}c\bigr)=c^{m}c^{S(j)},$$

so $S(j)\in E$. By Principle of Induction for the Natural Numbers, applied to $E$, we get $E=\mathbb{N}$; in particular $n\in E$.

Claim 7. Here $0\le1$ by claim 1 of Elementary Arithmetic in an Ordered Field, so $0\le c$ by transitivity, and clause 5 gives $0\le c^{k}$ for every $k\in\mathbb{N}$. We first show $1\le c^{k}$ for every $k\in\mathbb{N}$. Let $A$ be the set of $k\in\mathbb{N}$ with $1\le c^{k}$. Then $1\in A$, since $c^{1}=c$ by clause 1. Let $k\in A$. Claim 5 of Elementary Arithmetic in an Ordered Field, applied to $1\le c$ and $0\le c^{k}$, gives $c^{k}\cdot1\le c^{k}c$. Here $c^{k}c=c^{S(k)}$ by clause 1, and $c^{k}\cdot1=c^{k}$. Thus $1\le c^{k}=c^{k}\cdot1\le c^{k}c=c^{S(k)}$, and transitivity gives $1\le c^{S(k)}$, so $S(k)\in A$. By Principle of Induction for the Natural Numbers, applied to $A$, we get $A=\mathbb{N}$.

Now let $m\in\mathbb{N}$ with $m\le n$. If $m=n$, there is nothing to prove. Otherwise $m<n$, and claim 7 of Properties of the Order on the Natural Numbers gives $k\in\mathbb{N}$ with $n=m+k$. Clause 6, applied with the exponents $m$ and $k$, gives $c^{n}=c^{m}c^{k}$. Claim 5 of Elementary Arithmetic in an Ordered Field, applied to $1\le c^{k}$ and $0\le c^{m}$, gives $c^{m}=c^{m}\cdot1\le c^{m}c^{k}=c^{n}$.