# Proof involving the Euler phi-function

I'm having trouble approaching the following exercise:

Let $n$ and $k$ be positive integers. Prove that $\phi(n^k) = n^{k-1}\phi(n)$.

I've tried examining the prime factorization $n^k = (p_1^{a_1} \cdot\cdot\cdot p_r^{a_r})^k$ and plugging it into the Euler phi-function but I'm not really sure where to go from there.

Any hints?

The method you have suggested should work fine as long as you know the formula $$\phi(n)=(p_1-1)p_1^{a_1-1}\cdots(p_r-1)p_r^{a_r-1}\ .$$ Then we get $$n^k=p_1^{ka_1}\cdots p_r^{ka_r}\quad\Rightarrow\quad \phi(n^k)=(p_1-1)p_1^{ka_1-1}\cdots(p_r-1)p_r^{ka_r-1}\ ,$$ and I think I can leave the rest up to you.