Suppose $G$ is group of order $p^k$ for $p$ prime and $k$ positive integer. Prove $G$ must contain an element with an order of $p$.
I have proven that every element of $G$ must be of order $p^j$, $0\le j\le k$.
I was trying to put together a proof where some $g\in G$ generates a cyclic subgroup $H$. Since $H$ cyclic, then it's generated by $g$ from $G$ so $|g|=p^j$ for some $j$. So some element $h\in H$ is $h=g^r$, some $0\le r\le j$. And, $|h|=|g^r|= \frac{|g|}{gcd(r,|g|)}$, where $g$ is generator of $H$ so $|g|=p^j$. Then, taking $r=^j-1$, have $|h|=|p^j|/|p^{j-1}|=p$. So some $h\in H$ is $h\in G$ with $|h|=p$.