Power of Product of Commutative Elements in Monoid

Theorem
Let $\left ({S, \circ}\right)$ be a monoid whose identity is $e_S$.

Let $a, b \in S$ be invertible elements for $\odot$ that also commute.

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

Proof
From the similar result for semigroups, this holds if $n \ge 0$.

Since $a$ and $b$ commute, then so do $a^{-1}$ and $b^{-1}$ by Commutation of Inverses.

Hence, if $n > 0$:

The result follows.