Is this proof correct?
Prove that $$\sum_{d|n} \mu(d)\log^m(d) = 0$$ if $m \geq 1$ and $n$ has more than $m$ distinct prime factors.
By induction over the number $k$ of prime factors of $n$.
- Base case $k = 2$.
We have $m = 1$ and $\sum_{d|n} \mu(d)\log(d) = - \Lambda(n) = 0$
- Induction step
Suppose the statement holds for some $k > m$, and consider any $r \in \mathbb{Z}$ with $k+1$ prime factors. Then we write $r = n p^a$ where $p$ does not divide $n$. To account for all divisors of $r$ that contribute to our sum, we only need to consider all divisors of $n$ multiplied by either $1$ or $p$, giving:
$$\sum_{d|r} \mu(d)\log^m(d) = \sum_{d|n} \mu(d)\log^m(d) + \sum_{d|n} \mu(dp)\log^m(dp). $$
Since $n$ has $k > m$ prime factors, by our induction hypothesis, the first sum on the RHS is $0$. Also, since $\mu$ is multiplicative: \begin{align*}\sum_{d|n} \mu(dp)\log^m(dp) & = \sum_{d|n} \mu(d)\mu(p)\big(\log(d)+\log(p)\big)^m \\ & = - \sum_{d|n} \mu(d)\sum_{s=0}^m {m\choose s}\log^{m-s}(p)\log^s(d) \\ & = -\sum_{s=0}^m {m\choose s}\log^{m-s}(p) \sum_{d|n} \mu(d)\log^s(d).\end{align*}
Now, since $n$ has $k > m$ prime factors, by our induction hypothesis $\sum_{d|n} \mu(d)\log^s(d) = 0$ for all $0 \leq s \leq m$, and this concludes the proof.