Powers of Semigroup Element Commute

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

Let $a \in S$.

Let $m, n \in \N_{>0}$.

Then:
 * $\forall m, n \in \N_{>0}: a^n \odot a^m = a^m \odot a^n$

Proof
From Index Laws for Semigroup: Sum of Indices we have:


 * $\forall m, n \in \N_{>0}: a^{n+m} = a^n \odot a^m$

But from Natural Number Addition is Commutative, we have that $n+m = m+n$.

So:
 * $a^n \odot a^m = a^{n+m} = a^{m+n} = a^m \odot a^n$