Power of Product of Commutative Elements in Semigroup

Theorem
Let $\struct {S, \circ}$ be a semigroup.

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

Then:
 * $\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n \iff x \circ y = y \circ x$

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

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


 * $\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n$

Sufficient Condition
Suppose $\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n$.

In particular, when $n = 2$,