Let $G$ be a finite group, and let $k>1$ be an integer.  

I need to prove that if the mapping $f:G\rightarrow G$, defined by $f(g)=g^k$, is bijection, then $\gcd(k,|G|)=1$.  

I almost certain that if there exist such $f$, then $G$ is of prime order, and that will then give the answer (since $k\neq |G|$), I just couldn't find a way to prove it.  

Any help would greatly appreciated.