Element to Power of Remainder

Theorem
Let $$G$$ be a group whose identity is $$e$$.

Let $$a \in G$$ have finite order such that $$\left|{a}\right| = k$$.

Then:
 * $$\forall n \in \Z: n = q k + r: 0 \le r < k \implies a^n = a^r$$

Proof
Let $$n \in \Z$$.

Then from the Division Theorem, $$\exists q, r \in \Z: n = q k + r, 0 \le r < k$$.

From the index laws and the fact that $$a^k = e$$, we have:

$$ $$ $$ $$