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, $k \backslash n$, and so $\exists m \in \N^*: k m = n$.

Thus: