TITLE: Showing an indentity with a cyclic sum
QUESTION [8 upvotes]: Let  $n\geqslant2$, and $k\in \mathbb{N}$
Let $z_1,z_2,..,z_n$ be distinct complex numbers
Prove that
$$ \sum_{i=1}^{n}\frac  {{z}_{i}^{n-1+k}} { \prod \limits_{\substack{j = 1\\j \ne i}}^{ n }{ (z_i-z_j) } } =\large\sum_{i_1+i_2+...+i_n=k}^{ } {z}_{1}^{{i}_{1}} {z}_{2}^{{i}_{2}}...{z}_{n}^{{i}_{n}} $$

This originates from a problem involving Vandermonde 's determinant.
I've tried induction on $n$ and $k$ without success...

REPLY [12 votes]: The case $n=2$ is simple. If $z_i = 0$ for any $i$ then we can prove it by induction. So we suppose that $z_i \neq 0$ for each $i$. 
We choose $R > 0$ such that $|Rz_i| \leqslant 1$ for each $i$. Then $\displaystyle{ \sum_{m=0}^{\infty} (xz_i)^m}$ converges. We have 
$\displaystyle{ \prod_{i=1}^n \sum_{m=0}^{\infty} (xz_i)^m = \prod_{i=1}^n \frac{1}{1-xz_i} = \sum_{i=1}^n \frac{A_i}{1-xz_i}}$
for some $A_i$ (analysis in simple fractions). If $z_i = 0$ for any $i$ it is also $A_i = 0$. Otherwise multiplying both side by $(1-xz_i)$ and taking the limits we have 
$\displaystyle{ A_i = \lim_{x \to 1/z_i} \prod_{j \neq i} \frac{1}{1-xz_j} = \prod_{j \neq i} \frac{z_i^{n-1}}{z_i-z_j}.}$
So finally we have
$\displaystyle{\prod_{i=1}^n \sum_{m=0}^{\infty} (xz_i)^m = \sum_{i=1}^n \frac{z_i^{n-1}}{(1-xz_i)\prod_{j \neq i}(z_i-z_j)} =  \sum_{i=1}^n \frac{z_i^{n-1}}{\prod_{j \neq i}(z_i-z_j)}\sum_{m=0}^{\infty} (xz_i)^m.}$
Now comparing the coefficients of $x^k$ we get the result.