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

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

Let $\left({T, *}\right)$ be a semigroup, and let $a \in T$.

Let the mapping $*^n: S \to T$ be defined as:
 * $\forall n \in S: *^n a = \begin{cases}

e & : n = 0 \\ a & : n = 1 \\ \end{cases}$ where $e$ is the identity element of $\left({T, *}\right)$, if such exists.
 * ^r * a & : n = r \circ 1

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

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

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

From Uniqueness of Recursive Mapping from Naturally Ordered Semigroup to Algebraic Structure with Identity:
 * $*^n: S \to T$ is the only mapping with that definition.

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

Let $S'$ be the set of all $n \in S$ such that:


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

We have:
 * $*^0 \left({a * b}\right) = e = e * e = \left({*^0 a}\right) * \left({*^0 b}\right)$

So $0 \in S'$ whether or not $a * b = b * a$.

Now:

So $1 \in S'$.

Now suppose $n \in S'$. Then we have:

So $n \circ 1 \in S'$.

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