Showing an indentity with a cyclic sum

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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...