Powers of Commuting Elements of Semigroup Commute

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

That is:
 * $\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$

Proof
The proof proceeds by the Principle of Mathematical Induction:

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


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

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

This is the basis for the induction.

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


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

This is the induction hypothesis.

It remains to be shown that:
 * $P \left({k}\right) \implies P \left({k + 1}\right)$

That is, that:
 * $\left({\circ^{k + 1} a}\right) \circ b = b \circ \left({\circ^{k + 1} a}\right)$

Induction Step
This is the induction step:

Thus:

So $P \left({k + 1}\right)$ is true.

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

By repeating the argument above, replacing $a$ with $b$ and $b$ with $\circ^m a$, we have:

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

That is:
 * $\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$