Order of Finite p-Group is Power of p/Proof 2

Proof
Let every element of $G$ be a $p$-element.

Let $q$ be a prime number which is a divisor of the order $\order G$ of $G$.

By Cauchy's Lemma, there exists an element of $G$ whose order is a divisor of $q$.

But as the order of all elements of $G$ divide $p^n$ it follows that $q = p$.

Thus $G$ is a group whose order is $p^n$ for some $n \in \Z_{>0}$.