# Two Laurent polynomials actually being equal

I encountered two Laurent polynomials $$\sum_{j=0}^k\sum_{ \mbox{\begin{array}{c} i_1,\dots,i_k\geq 0\\ i_0=-N-i_1-\dots-i_k\end{array}}} \sum_{m=0}^{i_j} \begin{pmatrix} -N\\ i_1,\dots,i_k \end{pmatrix} \begin{pmatrix} i_j\\ m \end{pmatrix} x_0^{i_0}\cdots x^{i_j-m}\cdots x_k^{i_k}$$ and
$$\sum_{j=0}^k\sum_{N\geq p_1\geq\cdots\geq p_{n}\geq 0} \sum_{ \mbox{\begin{array}{c} i_1,\dots,i_k\geq 0\\ i_0=-N-i_1-\dots-i_k-p_1-\cdots- p_{n}\end{array}}} \begin{pmatrix} N\\ p_1 \end{pmatrix} \begin{pmatrix} p_1\\ p_2 \end{pmatrix}\cdots \begin{pmatrix} p_{n-1}\\ p_{n} \end{pmatrix} \begin{pmatrix} -(N+p_1+\cdots+p_n)\\ i_1,\dots,i_k \end{pmatrix} x_0^{i_0}\cdots x^{i_j+p_1+\cdots+p_n}\cdots x_k^{i_k},$$ where $N,n,k$ are given positive integers and $$\begin{pmatrix} l\\ i_1,\dots,i_k \end{pmatrix}= \begin{pmatrix} l\\ i_1 \end{pmatrix} \begin{pmatrix} l-i_1\\ i_2 \end{pmatrix}\cdots \begin{pmatrix} l-i_1-\cdots-i_{k-1}\\ i_k \end{pmatrix}$$ for any integer $l$, positive integers $i_1,\dots,i_k$.
I need to prove that they are actually the same.
Can anyone help me? or just give me some hint. I'll be grateful.

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Did you check that they are the same? (Say, by a computer experiment.) Accidentally, what are the parameters? $N$ and $k$? At first sight, your "polynomials" look more like power series. – Alex Degtyarev Mar 14 '14 at 12:30
There is no $i_0$, so $j$ should run in both sums from 1 to k. Using $\sum_{m=0}^{i} \binom imx^{i-m}=\sum_{m=0}^{i} \binom imx^{m}=(1+x)^i$ with $i=i_j$ repeatedly for each $j$, you should come up with some power of $(k+x)$ for both expressions. I wonder though if your summations are well-defined. – Wolfgang Mar 14 '14 at 18:09