Let G be a finite group with more than one element. Show that G has an element of prime order.
Let $\displaystyle a\in G, \ ord(a)=n$.
If $\displaystyle n$ is prime, we've done.
Else, let $\displaystyle p$ be a prime divisor of $\displaystyle n$.
Then $\displaystyle a^n=a^{pm}=(a^m)^p=e$.
Let $\displaystyle b=a^m\in G\Rightarrow b^p=e$.
Let $\displaystyle 0<q<p$. Then $\displaystyle b^q=(a^m)^q=a^{mq}$.
But $\displaystyle mq<mp=n\Rightarrow a^{mq}\neq e$.
So $\displaystyle ord(b)=p$.