How to prove $\sum_{ d \mid n} \mu(d)f(d)=\prod_{i=1}^r (1-f(p_i))$? I have to prove for $n \in \mathbb{N}>1$ with $n=\prod \limits_{i=1}^r p_i^{e_i}$. $f$ is a multiplicative function with $f(1)=1$:
$$\sum_{ d \mid n} \mu(d)f(d)=\prod_{i=1}^r (1-f(p_i))$$
How I have to start? Are there different cases or can I prove it in general? 
Any help would be fine :)

Please see Theorem 2.18 on page $37$ in Tom Apostol's Introduction to analytic number theory book.
The proof goes as follows:
Define $$ g(n) = \sum\limits_{d \mid n} \mu(d) \cdot f(d)$$

*

*Then $g$ is multiplicative, so to determine $g(n)$ it suffices to compute $g(p^a)$. But note that $$g(p^a) = \sum\limits_{d \mid p^{a}} \mu(d) \cdot f(d) = \mu(1)\cdot f(1) + \mu(p)\cdot f(p) = 1-f(p)$$
