Proof of Addition of Exponents for Natural Number Powers in a Field

Let $S$ be the successor map of Natural Numbers. By claim 1 of Properties of Natural Number Powers in a Field, the powers of $c$ satisfy $c^{1}=c$ and $c^{S(k)}=c^{k}c$ for every $k\in\mathbb{N}$.

Fix $m\in\mathbb{N}$ and let $E$ be the set of those $n\in\mathbb{N}$ for which $c^{m+n}=c^{m}c^{n}$.

Base. By claim 1 of Arithmetic of Addition on the Natural Numbers we have $m+1=S(m)$, so

$$c^{m+1}=c^{S(m)}=c^{m}c=c^{m}c^{1}.$$

Hence $1\in E$.

Induction step. Let $n\in E$. The recursive identity $m+S(n)=S(m+n)$ of Natural Numbers gives

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

where the second and last equalities are the recursion of claim 1 of Properties of Natural Number Powers in a Field, the third uses $n\in E$, and the fourth is associativity of multiplication in the field $K$. Hence $S(n)\in E$.

By Principle of Induction for the Natural Numbers, $E=\mathbb{N}$. Since $m\in\mathbb{N}$ was arbitrary, $c^{m+n}=c^{m}c^{n}$ for all $m,n\in\mathbb{N}$.