Index Laws/Product of Indices/Semigroup

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

Let $a \in S$.

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

Let $\circ^n \left({a}\right) = a^n$ be defined as the power of an element of a semigroup:


 * $a^n = \begin{cases}

a : & n = 1 \\ a^x \circ a : & n = x + 1 \end{cases}$

... that is:
 * $a^n = a \circ a \circ \cdots \left({n}\right) \cdots \circ a = \circ^n \left({a}\right)$

Then:
 * $\forall m, n \in \N_{>0}: a^{n m} = \left({a^n}\right)^m = \left({a^m}\right)^n$

Proof
This is proved in Naturally Ordered Semigroup Power Law.