Commutativity of Powers in Monoid

Theorem
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.

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

Then:
 * $\forall m, n \in \Z: a^m \circ b^n = b^n \circ a^m$

Proof
By Powers of Commuting Elements of Semigroup Commute, if $m > 0$ and $n > 0$ then $a^m$ commutes with $b^n$.

By Commutation with Inverse in Monoid, again if $m > 0$ and $n > 0$ then $a^m$ commutes with $\paren {b^n}^{-1} = b^{-n}$.

Similarly $b^n$ commutes with $a^{-m}$.

But as $a^{-m}$ commutes with $b^n$, it also commutes with $\paren {b^n}^{-1} = b^{-n}$, again by Commutation with Inverse in Monoid.

Hence the result.