This is a really classical result, but this proof is just so magical that I have to include it.
Lemma. Let be a finite group, and let be a prime dividing . Then has an element of order .
Proof. Consider the action of on
Then , since the orbits of size all have size (and the fixed points are exactly the union of the orbits of size 1). The right hand side is as . Thus,
But there is a bijection
The former trivially contains the element 1, hence (since its size is divisible by ) it has to contain some other element . This element will then have (exact) order .
Remark. Of course, the result also follows from the much stronger Sylow theorems (of which we only need the existence statement).