Power of Product of Commutative Elements in Semigroup

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

Let $x, y \in S$ both be cancellable elements of $S$.

Then:
 * $\forall n \in \N_{>1}: \left({x \circ y}\right)^n = x^n \circ y^n \iff x \circ y = y \circ x$

Necessary Condition
Let $x \circ y = y \circ x$.

Then by Power of Product of Commuting Elements in Semigroup equals Product of Powers:


 * $\forall n \in \N_{>1}: \left({x \circ y}\right)^n = x^n \circ y^n$

Proof 1
As $x \circ y^n = y^n \circ x$ if and only if $x$ and $y$ commute, the result follows.

Proof 2
Suppose $\forall n \in \N_{>1}: \left({x \circ y}\right)^n = x^n \circ y^n$.

In particular, when $n = 2$,