Index Laws/Sum of Indices/Semigroup

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

Let $a \in S$.

Let $n \in \N^*$.

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


 * $a^n = \begin{cases}

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

... that is:
 * $a^n = \underbrace{a \circ a \circ \cdots \circ a}_{n \text{ copies of } a} = \circ^n \left({a}\right)$

Then:
 * $\forall m, n \in \N^*: a^{n+m} = a^n \circ a^m$

Proof
Because $\left({S, \circ}\right)$ is a semigroup, $\circ$ is associative on $S$.

Let $T$ be the set of all $m \in \N$ such that this result holds.

Let $n \in \N^*$.

So $1 \in T$.

Now suppose $m \in T$. Then we have:

So $m + 1 \in T$.

So by the Principle of Finite Induction, $T = \N^*$.

Thus this result is true for all $m, n \in \N^*$.