This is exercise $2.35$ from Rotman's A First Course in Abstract Algebra.
Let $G$ be a group and let a $a \in G$ have order $pk$ for some prime $p$, where $k \geq 1$. Prove that if there is $x \in G$ with $x^p = a$, then the order of $x$ is $p^2k$, and hence $x$ has larger order than $a$.
This isn't homework, but I'm stuck. Is there a nice way to prove it?