Prove that for every positive integer $\displaystyle n$, $\displaystyle \sum_{d|n}\mu(d)d(d)=(-1)^{\omega(n)}$.

The notation here is $\displaystyle \mu$ denoting Mobius inversion, with

$\displaystyle \omega(n)=\sum_{p|n}1$ (the number of distinct primes dividing $\displaystyle n$), and

$\displaystyle d(n)=\sum_{d|n}1=\prod_{p^{\alpha}\parallel n}(\alpha+1)$ (the number of positive divisors of $\displaystyle n$)

I'm pretty stuck on this one. Any help would be much appreciated!