For whatever reason, I am having trouble proving the following claim without using LaGrange's Theorem.
Claim: Let $G$ be a group of order $n < \infty$. Then, $x^{n}=1$, where $1$ is the identity, for all $x \in G$.
If $|x|=m$, then I can use the Division Algorithm (specifically, I can divide $n$ by $m$) to obtain that $x^{n} = x^{r}$, where $r=0,1,2,..., \text{or}\> m-1$. Then, well, I can't see the next step.