Let $\mu$ be the Mobius function, and $\phi$ Euler's totient function. I am reading a proof found in this paper (Theorem 2 on page 17), and I can't quite figure why I'm getting something different from what they have in equation (18): For an integer $k \geq 2$, \begin{align*} \sum_{a = 1}^{\infty} \frac{\mu(a)}{\phi(a^k)} = \prod_{p \text{ prime}} \left(1 - \frac{1}{p^{k-1}(p-1)}\right). \end{align*}
Here is how I calculate the series as an Euler product: I can make the series on the left a Dirichlet series using the identity $\phi(a^k) = a^k \prod_{p \mid a} \left(1 - 1/p\right)$, so \begin{align*} \sum_{a = 1}^{\infty} \frac{\mu(a)}{\phi(a^k)} = \sum_{a = 1}^{\infty} \frac{f(a)}{a^k} \end{align*} where $f(a) = \mu(a)/\prod_{p \mid a} \left(1 - 1/p\right)$, (which I'm pretty sure is multiplicative). Then we have the Euler product \begin{align*} \sum_{a = 1}^{\infty} \frac{f(a)}{a^k} = \prod_{p} P(p) \end{align*} where \begin{align*} P(p) = \sum_{n \geq 0} \frac{f(p^n)}{p^{nk}} = 1 - \frac{1}{p^k(p-1)}. \end{align*}
I think this is right, but why am I getting $P(p) = 1 - 1/\phi(p^{k+1})$ instead of $1 - 1/\phi(p^{k})$?
Well, $f(p^0)=1$, $f(p^1)=-1/(1-1/p)=-p/(p-1)$, and $f(p^n)=0$ for $n>1$, thus $$P(p)=1-\frac1{p^k}\frac{p}{p-1}=1-\frac1{p^{k-1}(p-1)}.$$