Does $\phi(n^k) = {n^{k-1}}\phi(n)$ always hold?

Question

Let $\phi$ be Euler's totient function.

Here is my question:

Does $\phi(n^k) = {n^{k-1}}\phi(n)$ always hold, where $k \in \mathbb{N}$?

I came across this claim as (A. Olofsson, pers. comm., Dec. 30, 2004) from the MathWorld - Totient Function page.

Answer 1

Yes, it's true.

Let $n,k$ be positive integers.

Let $p_1,...,p_s$ be the distinct prime factors of $n$.

Then we have the formula $$\phi(n) = n(1 - 1/p_1)\cdots (1-1/p_s)\qquad\,$$ But $n$ and $n^k$ have the same distinct prime factors, hence \begin{align*} \phi(n^k) &= n^k(1 - 1/p_1)\cdots (1-1/p_s)\\[4pt] &= n^{k-1}(n(1 - 1/p_1)\cdots (1-1/p_s))\\[4pt] &= n^{k-1}\phi(n) \end{align*}