# Identity Involving $\sum_{d|n} \mu(d) \log^m(d)$

The problem is: prove that $$\sum_{d|n} \mu(d) \log^m(d) = 0$$ if $m \ge 1$ and $n$ has more than $m$ distinct prime factors. Here, $\mu$ is the Möbius function.

This is from a book I'm self studying from, and it hints to use induction. I prove the identity for $m = 1$ by using the fact that $$\Lambda(n) = -\sum_{d|n} \mu(d) \log(d)$$ and the definition of the Mangoldt function, $\Lambda$. I then fiddled around with the sum for a bit to try to prove the inductive step, but I haven't managed to get anywhere useful.

Hints would be preferable to answers, but I'm happy to have either.

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Hint: Let $p$ be a prime factor of $n$. Write $n=p^as$, where $a \geq 1, p \nmid s$. Split up the sum into $$\sum_{d \mid s}{\mu(d)\log^m(d)}+\sum_{d\mid s}{\mu(pd)\log^m(pd)}$$ and recombine to get a sum over $d \mid s$, where $s$ has $>m-1$ prime factors. You still need to do some work before you can apply the induction hypothesis.