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

Proof
Let $P \left({n}\right)$ be the proposition that $\left({x \circ y}\right)^n = x^n \circ y^n$ iff $x$ and $y$ commute.

We note in passing that $P \left({1}\right)$ does not hold: it says $x \circ y = x \circ y \iff x \circ y = y \circ x$ which is just wrong.

Next we note that $P \left({2}\right)$ holds, as follows:

Now suppose $P \left({n}\right)$ holds. We need to show that $P \left({n+1}\right)$ holds as a result of this.

That is, that $\left({x \circ y}\right)^{n+1} = x^{n+1} \circ y^{n+1} \iff x \circ y = y \circ x$.

Suppose $x \circ y = y \circ x$ commute. Then:

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