Showing an indentity with a cyclic sum 
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...
 A: I would say here thart the difficulty is more in the notations than in the mathematics. I present below
a proof by induction on $n$, where the result for $\lbrace z_1,z_2,\ldots,z_n \rbrace$ is deduced
from the result for $\lbrace z_1,z_2,\ldots,z_{n-1} \rbrace$ and the result for 
$\lbrace z_1,z_2,\ldots,z_{n-2},z_n \rbrace$. This proof, although completely elementary and
natural, is full of lengthy formulas on several lines, and I suspect that there are shorter
and more elegant proofs.
To ease notation a little, let us put 
$$
M(n,\lbrace x_1,x_2,\ldots,x_r \rbrace)=
\sum_{l_1+l_2+\ldots+l_r}x_1^{l_1}x_2^{l_2}\ldots x_r^{l_r}, \
P(y,\lbrace x_1,x_2,\ldots,x_r \rbrace)=
\prod_{k=1}^r\frac{1}{y-x_k} \tag{1}
$$
Your identity can then be reformulated as
$$
\sum_{i=1}^n z_i^{k+n-1}P(z_i,\lbrace z_j\rbrace_{j\neq i})=
M(k,\lbrace z_1,z_2,\ldots,z_n \rbrace) \tag{2}
$$
Throughout this proof, we shall use freely the following identities,
which follow immediately from the definitions in
(1) :
$$
M(n,X)=\sum_{t=0}^n M(n-t,X\setminus\lbrace x \rbrace)x^t, \ \ \ 
P(y,X)=\frac{1}{y-x}P(y,X\setminus\lbrace x \rbrace). \tag{3} 
$$
We prove (2) by induction on $n\geq 2$. When $n=2$, (2) reduces
to a familiar and easy identity in $\frac{z_2}{z_1}$. So let us assume
$n>2$ and that (2) holds for every $n'<n$. First ,using (2) on
the $n-1$-element set $\lbrace z_1,z_2,\ldots,z_{n-2},z_n\rbrace$,
we have
$$
\Bigg(\sum_{i=1}^{n-2} z_i^{k+n-1}P(z_i,\lbrace z_j\rbrace_{j\neq i,j\neq n-1})\Bigg)+
z_n^{k+n-1}P(z_n,\lbrace z_j\rbrace_{j\leq n-2})=
M(k+1,\lbrace z_1,z_2,\ldots,z_{n-2},z_n \rbrace) \tag{4}
$$
Or in other words,
$$
\Bigg(\sum_{i=1}^{n-2} \frac{z_i^{k+n-1}}{z_i-z_n}P(z_i,\lbrace z_j\rbrace_{j\neq i,j\leq n-2})\Bigg)+
z_n^{k+n-1}P(z_n,\lbrace z_j\rbrace_{j\leq n-2})=
\sum_{t=0}^{k+1}M(k+1-t,\lbrace z_1,z_2,\ldots,z_{n-2} \rbrace)z_n^t \tag{5}
$$
Divide this by $z_n-z_{n-1}$ :
$$
\Bigg(\sum_{i=1}^{n-2} \frac{z_i^{k+n-1}}{(z_i-z_n)(z_n-z_{n-1})}
P(z_i,\lbrace z_j\rbrace_{j\neq i,j\leq n-2})\Bigg)+
z_n^{k+n-1}P(z_n,\lbrace z_j\rbrace_{j\leq n-1})=
\sum_{t=0}^{k+1}M(k+1-t,\lbrace z_1,z_2,\ldots,z_{n-2} \rbrace)\frac{z_n^t}{z_n-z_{n-1}} \tag{6}
$$
Now, each of the terms in (6) can be expanded in its own way : we have
$$
\begin{array}{lcl}
\frac{1}{(z_i-z_n)(z_n-z_{n-1})} &=& 
\frac{1}{(z_i-z_{n-1})(z_n-z_{n-1})}+\frac{1}{(z_n-z_{i})(z_{n-1}-z_i)}\\
\frac{z_n^t}{z_n-z_{n-1}} &=& 
\sum_{s=0}^{t-1}z_{n-1}^{t-1-s}z_n^s+\frac{z_{n-1}^t}{z_n-z_{n-1}}\\
\end{array}\tag{7}
$$
whence
$$
\begin{array}{l}
\frac{z_i^{k+n-1}P(z_i,\lbrace z_j\rbrace_{j\neq i,j\leq n-2})}{(z_i-z_n)(z_n-z_{n-1})} \\ =
\frac{z_i^{k+n-1}P(z_i,\lbrace z_j\rbrace_{j\neq i,j\leq n-2})}{z_n-z_{n-1}}+z_i^{k+n-1}
P(z_i,\lbrace z_j\rbrace_{j\neq i})\\
\text{and} \\
\sum_{t=0}^{k+1}M(k+1-t,\lbrace z_1,z_2,\ldots,z_{n-2} \rbrace)\frac{z_n^t}{z_n-z_{n-1}} \\ =
M(k,\lbrace z_1,z_2,\ldots,z_n \rbrace)+\frac{M(k+1,\lbrace z_1,z_2,\ldots,z_{n-1} \rbrace)}{z_n-z_{n-1}}\\
\end{array}\tag{8}
$$
Injecting (8) into (6), we deduce
$$
\begin{array}{l}
\Bigg(\sum_{i=1}^{n-2} \frac{z_i^{k+n-1}}{z_n-z_{n-1}}P(z_i,\lbrace z_j\rbrace_{j\neq i,j\leq n-2})\\
+\sum_{i=1}^{n-2} z_i^{k+n-1}P(z_i,\lbrace z_j\rbrace_{j\neq i})\Bigg) \\
+z_n^{k+n-1}P(z_n,\lbrace z_j\rbrace_{j\leq n-1})\\
=M(k,\lbrace z_1,z_2,\ldots,z_n \rbrace)+
\frac{M(k+1,\lbrace z_1,z_2,\ldots,z_{n-1} \rbrace)}{z_n-z_{n-1}} 
\end{array} \tag{9}
$$
In other words,
$$
\begin{array}{l}
\Bigg(\sum_{i\neq n-1}z_i^{k+n-1}P(z_i,\lbrace z_j\rbrace_{j\neq i})\Bigg)
-M(k,\lbrace z_1,z_2,\ldots,z_n \rbrace) \\ = \\
+\frac{M(k+1,\lbrace z_1,z_2,\ldots,z_{n-1} \rbrace)}{z_n-z_{n-1}}-
\Bigg(\sum_{i=1}^{n-2} \frac{z_i^{k+n-1}}{z_{n}-z_{n-1}}P(z_i,\lbrace z_j\rbrace_{j\neq i,j\leq n-2})\Bigg) 
\end{array} \tag{10}
$$
Now, adding $z_{n-1}^{k+n-1}P(z_{n-1},\lbrace z_j\rbrace_{j\neq n-1})=
\frac{z_{n-1}^{k+n-1}P(z_{n-1},\lbrace z_j\rbrace_{j\neq n-1,j\neq n})}{z_{n-1}-z_{n}}$ to both sides of this equality,
we obtain
$$
\begin{array}{l}
\Bigg(\sum_{i}z_i^{k+n-1}P(z_i,\lbrace z_j\rbrace_{j\neq i})\Bigg)
-M(k,\lbrace z_1,z_2,\ldots,z_n \rbrace) \\ = \\
+\frac{M(k+1,\lbrace z_1,z_2,\ldots,z_{n-1} \rbrace)-
\sum_{i=1}^{n-1}z_i^{k+n-1}P(z_i,\lbrace z_j\rbrace_{j\neq i,j\leq n-1})}{z_n-z_{n-1}}
\end{array} \tag{11}
$$
The RHS must be zero by the induction hypothesis, which concludes the proof.
A: 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.
