Element to Power of Group Order is Identity

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

Then:
 * $\forall g \in G: g^n = e$

Proof
Let $G$ be a group such that $\left|{G}\right| = n$.

Let $g \in G$ and let $\left|{g}\right| = k$.

From Order of Element Divides Order of Finite Group:
 * $k \mathop \backslash n$

So:
 * $\exists m \in \Z_{>0}: k m = n$

Thus: