A sufficient condition for a finite group to have a $p$-power order

Question

Let $G$ be a finite group and $p$ a prime number. If $G$ is a $p$-group, i.e. the order of every element of $G$ is a power of $p$ then is the order of $G$ equal to some power of $p$? How do I show this?

Answer 1

By Cauchy's theorem, if there is another $q$ dividing $|G|$, then there exist an element of order $q$ in $G$ for prime $q$ which conclude the result.

Answer 2

If $|G|$ has another prime-power factor $q^k$ ($k$ maximal), by the first Sylow's theorem, it has a subgroup whose order is precisely $q^k$. Then, if every subgroup has an order $p^j$, the order of $G$ can not have divisors that are not powers of $p$.