Finite sum: $\sum_{i=1}^n\prod_{j=1,j\neq i}^n\frac{\alpha+z_i-z_j}{z_i-z_j}=n$ Any ideas of how to prove:
$\sum_{i=1}^n\prod_{j=1,j\neq i}^n\frac{\alpha+z_i-z_j}{z_i-z_j}=n$
where $\alpha$ is a constant. Assume $z_i\neq z_j$.

For $\alpha=0$, this is clear.
Fix $\alpha\not =0$ and put $P(x)=\prod (x-z_i)$ and $Q(x)=P(x+\alpha)$. Decompose $F(x)=\frac{Q(x)}{P(x)}$ in simple fractions:
$$F(x)=1+\sum\frac{A_i}{x-z_i}$$
We have $$ A_i=\frac{Q(z_i)}{\prod_{j\not =i}(z_i-z_j)}=\alpha \prod_{j\not =i}\frac{\alpha+z_i-z_j}{z_i-z_j}$$
Hence 
$$\frac{P(x+\alpha)-P(x)}{P(x)}=\alpha\sum_{i=1}^{n} \prod_{j\not =i}\frac{(\alpha+z_i-z_j)}{z_i-z_j}\frac{1}{x-z_i}$$
Now multiply by $x$, and let $x\to \infty$, it is easy to finish.
