Index Laws/Sum 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:
 * $a^n = \underbrace{a \odot a \odot \cdots \odot a}_{n \text{ copies of } a} = \odot^n \left({a}\right)$

Let $a \in S$. Then:


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

Proof
Because $\left({S, \odot}\right)$ is a semigroup, $\odot$ 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 $m, n \in \N^*$.