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 $\circ$ that also commute.

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

Proof
From Power of Product of Commutative Elements in Semigroup, this result 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.