Show $\sum_{c|n} \mu(c)f(c) = \{1-f(p_1)\}\{1-f(p_2)\} \dots \{1-f(p_r)\}$

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$n=p^{k_1}_1p^{k_2}_3...p^{k_r}_r$ and f is multiplicative function.I have tried convolution but it seems not solving.

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$$\sum_{c|n}\mu(c)f(c)=\prod_{i=1}^r\left(\sum_{p_i|n, q=0}^{k_r}\mu(p_i^q)f(p_i^q)\right)$$

But $$\sum_{p_i|n, q=0}^{k_r}\mu(p_i^q)f(p_i^q)=f(1)-f(p_i)$$