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 $c\,d$. All references to numbered claims below 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

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 $c^{n}=\prod_{k=1}^{n}a_{k}\le\prod_{k=1}^{n}b_{k}=d^{n}$.