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:

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