Index Laws/Product of Indices/Semigroup

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

Let $a \in S$.

Let $n \in \N^*$.

Let $\odot^n \left({a}\right) = a^n$ be defined as in Power of an Element:


 * $a^n = \begin{cases}

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

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

Let $a \in S$. Then:


 * $\forall m, n \in \N^*: a^{n m} = \left({a^n}\right)^m = \left({a^m}\right)^n$

Proof
This is proved in Naturally Ordered Semigroup Power Law.