Index Laws/Sum of Indices/Monoid

Theorem
Let $\left({S, \circ}\right)$ be a monoid whose identity element is $e$.

For $a \in S$, let $\circ^n a = a^n$ be defined as the $n$th power of $a$:


 * $a^n = \begin{cases}

e & : n = 0 \\ a^x \circ a & : n = x + 1 \end{cases}$

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

while:
 * $a^0 = e$

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

Proof
Because $\left({S, \circ}\right)$ is a monoid, it is a fortiori also a semigroup.

From Index Laws for Semigroup: Sum of Indices:


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

That is:
 * $\forall m, n \in \N_{>0}: a^{n + m} = a^n \circ a^m$

It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$.

Let $n \in \N$:

Similarly, let $m \in \N$:

and:

Thus:
 * $a^{n + m} = a^n \circ a^m$

holds for $n = 0$ and $m = 0$.

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

Also see

 * Power of Identity is Identity