Prove that $$\sum_{d|n}\mu(d)\log^m d=0$$ if $m\ge 1$ and $n$ has more than $m$ distinct prime factors.
I tried using Induction and kind of succeeded in the sense that if we write down the case $m=2$ and $n=pqr$, $p,q,r$ being distinct primes, the general case becomes quite evident- \begin{align*} \sum_{d|pqr}\mu(d)\log^2 d &= \mu(1)\log^2 1+\mu(p)\log^2 p+\mu(q)\log^2 q+\mu(r)\log^2 r+\mu(pq)\log^2 pq+\mu(qr)\log^2 qr+\mu(pr)\log^2 pr+\mu(pqr)\log^2 pqr\\ &=0-\log^2 p-\log^2 q-\log^2 r+(\log p+\log q)^2+(\log q+\log r)^2+(\log p+\log r)^2-(\log p+\log q+\log r)^2 \end{align*} from which it is intuitively clear that the terms will cancel out to give $0$.
But, to use Induction, we need to break a $\log^{m+1}d$ into lower powers of $\log d$, and the cancellation will be very difficult to show (especially considering the fact that it would require multinomial theorem).
So, is there an easy way out? Can we somehow use Dirichlet Multiplication or Möbius Inversion Formula?
A duplicate question has been suggested, but the answer here ends with "You still need to do some work before you can apply the induction hypothesis" which is precisely where I am having my troubles.
A way to finish the claim of the linked post is the following:
The required sum $$\sum_{d \mid s}{\mu(d)\log^m(d)}+\sum_{d\mid s}{\mu(pd)\log^m(pd)}$$ can be combined as $$\sum_{d \mid s} \left[ {\mu(d)\log^m(d)+\mu(p)\mu(d)(\log (p) + \log(d))^m} \right].$$ Using the binomial theorem and the fact that $(d,p)=1$, $$\sum_{d \mid s}{ \mu(d)\left[ \log^m(d)- \left(\log^m(d) + \sum_{0\leq k \leq m-1}{ \binom{m}{k} \log^{k}(d)} \log^{m-k}(p)\right) \right]}.$$ which simplifies to
$$-\sum_{d \mid s}{ \mu(d)\left[ \sum_{0\leq k \leq m-1}{ \binom{m}{k} \log^{k}(d)} \log^{m-k}(p) \right]}.$$ Once we apply the fundamental theorem of analytic number theory (aka switching sums), we get $$-\sum_{0\leq k \leq m-1}\binom{m}{k}\log^{m-k}(p)\sum_{d \mid s}{ \mu(d) { \log^{k}(d)} }.$$
The inner sum is $0$ by the induction hypothesis for all $k$ in the given range since the number of prime factors of $s$ is $>m-1 \geq k$.