Power of Product of Commuting Elements in Semigroup equals Product of Powers

Theorem
Let $\left({S, \circ}\right)$ be a semigroup.

For $a \in S$, let $\circ^n a = a^n$ denote the $n$th power of $a$.

Let $a, b \in S$ such that $a$ commutes with $b$:
 * $a \circ b = b \circ a$

Then:
 * $\forall n \in \N_{>0}: \circ^n \left({a \circ b}\right) = \left({\circ^n a}\right) \circ \left({\circ^n b}\right)$

That is:
 * $\forall n \in \N_{>0}: \left({a \circ b}\right)^n = a^n \circ b^n$

Proof
The proof proceeds by the Principle of Mathematical Induction:

Let $P \left({n}\right)$ be the proposition:


 * $\circ^n \left({a \circ b}\right) = \left({\circ^n a}\right) \circ \left({\circ^n b}\right)$

Basis of the Induction
So $P \left({1}\right)$ is true.

This is the basis for the induction.

Induction Hypothesis
Suppose that $P \left({k}\right)$ holds:


 * $\circ^k \left({a \circ b}\right) = \left({\circ^k a}\right) \circ \left({\circ^k b}\right)$

This is the induction hypothesis.

It remains to be shown that:


 * $P \left({k}\right) \implies P \left({k + 1}\right)$

That is, that:


 * $\circ^{k + 1} \left({a \circ b}\right) = \left({\circ^{k + 1} a}\right) \circ \left({\circ^{k + 1} b}\right)$

Induction Step
So $P \left({k + 1}\right)$ holds.

Thus by the Principle of Mathematical Induction, the result holds for all $n \in \N_{>0}$:


 * $\forall n \in \N_{>0}: \circ^n \left({a \circ b}\right) = \left({\circ^n a}\right) \circ \left({\circ^n b}\right)$