Element of Finite Group is of Finite Order/Proof 1

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

From Element has Idempotent Power in Finite Semigroup, for every element in a finite semigroup, there is a power of that element which is idempotent.

As $G$, being a group, is also a semigroup, the same applies to $G$.

That is:
 * $\forall x \in G: \exists n \in \N_{>0}: x^n \circ x^n = x^n$

From Identity is only Idempotent Element in Group, it follows that:
 * $x^n \circ x^n = x^n \implies x^n = e$

So $x$ has finite order.