A finite abelian group has order $p^n$, where $p$ is prime, if and only if the order of every element of $G$ is a power of $p$ Suppose that G is a finite Abelian group. Prove that G has order $p^n$, where p is prime, if and only if the order of every element of G is a power of p.

I tried the following route, but got stuck.  Using the fundamental theorem of finite Abelian groups, the problem reduces to proving Cauchy's theorem for a cyclic abelian group. If G is a cyclic group, and p divides G, then G has an element of order p whether p is prime or not. If we regard G as the integers mod p, then we can notice that if $|G| = kp$ then the integer k has order p in G. 

See: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cauchyapp.pdf 
To be exact, see theorem 1.1, it relies on Cauchy's Theorem. As already pointed out, the "abelian" part is not necessary.
