Finite Group is p-Group iff Order is Power of p

Theorem
Let $p$ be a prime number.

Definition 1 implies Definition 2
Let $G$ be a $p$-group by definition 1.

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

Let $g \in G$.

From Order of Element Divides Order of Finite Group, the order of $g$ is a divisor of $p^n$.

That is, $x$ is a $p$-element by definition.

As $x$ is arbitrary, it follows that all elements of $G$ are $p$-elements.

Thus $G$ is a $p$-group by definition 2.

Definition 2 implies Definition 1
Let $G$ be a $p$-group by definition 2.

Thus every element of $G$ is a $p$-element.

Let $q$ be a prime number which is a divisor of the order $\left|{G}\right|$ of $G$.

By Cauchy's Group Theorem, 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 $n \in \Z_{>0}$.

Thus $G$ is a $p$-group by definition 1.