Suppose $P$ is a finite poset, and $f: P \to \mathbb{C}$. Is it true that
$$\sum_{x_1<x_2<...<x_k}{(f(x_1)-1)(f(x_2)-1)(f(x_3)-1)...(f(x_k)-1)} = \sum_{x_1<x_2<...<x_k}(-1)^k \mu(0,x_1) \mu(x_1,x_2) \mu(x_2,x_3)...\mu(x_k,1)f(x_1)f(x_2)...f(x_k)$$
where the counter go to each chain $x_1<x_2<...<x_k $ of length $k$, for each $k \le$ poset's grade?
I have got problem finding the proof, as I only know that I should use the fact $\mu_k (0,1)= \sum_{k=1}^n (-1)^{(i+1)}N_i$ , where $n$ is the poset's grade, and $N_i$ is the number of chains of length $i$.