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

Theorem
Let $\left({T, \oplus}\right)$ be a semigroup.

For $a \in T$, let $\oplus^n a = a^n$ be defined as the $n$th power of $a$:


 * $a^n = \begin{cases}

a & : n = 1 \\ a^x \oplus a & : n = x + 1 \end{cases}$

That is:
 * $a^n = \underbrace{a \oplus a \oplus \cdots \oplus a}_{n \text{ copies of } a} = \oplus^n \left({a}\right)$

Let $a, b \in T$ such that $a$ commutes with $b$:
 * $a \oplus b = b \oplus a$

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

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

Proof
Let $a, b \in T: a \oplus b = b \oplus a$.

Because $\left({T, \oplus}\right)$ is a semigroup, $\oplus$ is associative on $T$.

By definition of $\oplus^n a = a^n$:
 * $\oplus^n a = \begin{cases}

a & : n = 1 \\ \left({\oplus^x a}\right) \oplus a & : n = x + 1 \end{cases}$

The proof proceeds by the Principle of Finite Induction:

Let $S$ be the set of all $n \in \N_{>0}$ such that:


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

Basis of the Induction
So $1 \in S$.

This is the basis for the induction.

Induction Hypothesis
Suppose that $k \in S$.

That is:


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

This is the induction hypothesis.

It remains to be shown that:
 * $k \in S \implies k + 1 \in S$

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

Induction Step
So $n + 1 \in S$.

Thus by the Principle of Finite Induction, $S = \N_{>0}$, and the result holds for all $n \in \N_{>0}$:
 * $\forall n \in \N_{>0}: \oplus^n \left({a \oplus b}\right) = \left({\oplus^n a}\right) \oplus \left({\oplus^n b}\right)$