Powers of Group Element Commute

Theorem
Let $\left ({G, \circ}\right)$ be a group.

Let $g \in G$.

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

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

Proof
By definition, a group is also a semigroup.

The result follows as a special case of Powers of Semigroup Element Commute