I need some help with the last(?) step in a proof and I'm not sure how I should proceed... $\mu(n)$ is the Möbius function and $\omega(n)$ is the number of distinct prime factors. We see that for $n$ prime, $\sum_{d \mid n}\mu(d)\omega(n/d )=\mu(1)\omega(n)+\mu(n)\omega(1)=1.$ But for $n$ composite I'm at a loss. My idea is to somehow show that if $n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$, then the terms in the following sum will cancel each other, like this: $\sum_{d \mid n}\mu(d)\omega(n/d )=\omega(n)-\sum\omega\left(\frac{n}{p_i}\right)+\sum\omega\left(\frac{n}{p_ip_j}\right)+\ldots+\sum\omega\left(\frac{n}{p_1p_2\ldots p_k}\right)(-1)^{k},$ where we just skip the divisors with exponent $a_i>1$ since $\mu$ would cancel them. Is there a way to achieve this? Or is this nonsense?
Added (clarification):
For $n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$, $\sum_{d \mid n}\mu(d)\omega \left( \frac{n}{d} \right)=\omega(n)+\underbrace{\mu(p_1) \omega(n/p_1)+ \ldots+\mu(p_k) \omega(n/p_k)}_{-\sum\omega\left(\frac{n}{p_i}\right)}+ \ldots + \mu(p_1 \ldots p_k) \omega\left(\frac{n}{p_1\ldots p_k}\right)$ where $\mu(p_i)=-1$ and $\mu(p_1 \ldots p_k)=(-1)^{k}$. So we don't sum terms with square factors since they would be 0 in any case.