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

Theorem
Let $\struct {S, \circ}$ 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:
 * $\forall n \in \N_{>0}: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$

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

Proof
The proof proceeds by the Principle of Mathematical Induction:

Let $\map P n$ be the proposition:


 * $\circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$

Basis of the Induction
So $\map P 1$ holds.

This is the basis for the induction.

Induction Hypothesis
Suppose that $\map P k$ holds:


 * $\map {\circ^k} {a \circ b} = \paren {\circ^k a} \circ \paren {\circ^k b}$

This is the induction hypothesis.

It remains to be shown that:


 * $\map P k \implies \map P {k + 1}$

That is, that:


 * $\map {\circ^{k + 1} } {a \circ b} = \paren {\circ^{k + 1} a} \circ \paren {\circ^{k + 1} b}$

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

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


 * $\forall n \in \N_{>0}: \map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$