If a group is finite, then every element has finite order.

Proof

Suppose $G$ is a finite group. Let $g \in G$ .
Consider ${g, g^{2}, \dots, g^{n}, \dots} = {g^{n} : n \in N}$ which is necessarily finite, contained in $G$ .
(This is because $G$ is necessarily closed under the operation.)

By pigeonhole principle, there exists some $m > n$ with $g^{m} = g^{n}$ .
Therefore,

\begin{aligned} g^{m - n} \cdot g^{n} & = g^{n} \\ g^{m - n} & = e \end{aligned} $ $ t h u s $ | g | \leq m - n < \infty $ . Q E D