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I am trying to prove the following identity for a set of variable $\lbrace z_1,\dots,z_k\rbrace$, for $S_k$ the set of all permutations of $[1,k]$ and $(-1)^P$ the signature of a permutation P. $$\sum_{P\in S_k} (-1)^P \prod_{1\leq i

There is a simple proof using the fact that the left hand side is an antisymmetric polynomial in the set of variable, and therefore it should be divisible by the Vandermonde determinant.

Yet I would like to do the full calculation, and I have started doing an induction. Supposing the relation is true for $k\in \mathbb{N}^*$ and calling $f(z_1,\dots,z_{k+1})=\sum_{P\in S_{k+1}}(-1)^P\prod_{1\leq i

$$f(z_1,\dots,z_{k+1})=\sum_{l=1}^{k+1}\prod_{\substack{1\leq i\leq k+1\\i \neq l}}(z_{i}-z_{l}-1) \sum_{\substack{P\in S_{k+1}\\ P(k+1)=l}} (-1)^P\prod_{1\leq i

I then define the permutation $\tilde{P}$ with the following property \begin{equation} \begin{cases} \tilde{P}(l)=k+1\\ \tilde{P}(k+1)=l\\ \mathrm{else}\; \tilde P(i)=i \\ \end{cases} \end{equation}

The signature of this permutation is $(-1)^{k+1-l}$ ; I then use the fact that a permutation $P$ of $S_{k+1}$ with $P(k+1)=l$ can be seen as a "composition" of a permutation $P'$ in $S_k$ with $\tilde P$, therefore I can rewrite the last part of my expression to use my induction hypothesis : $$f(z_1,\dots,z_{k+1})=\sum_{l=1}^{k+1}(-1)^{k+1-l}\prod_{\substack{1\leq i\leq k+1\\i \neq l}}(z_{i}-z_{l}-1) \sum_{P'\in S_{k}} (-1)^{P'}\prod_{1\leq i

This yields : $$f(z_1,\dots,z_{k+1})=k! \sum_{l=1}^{k+1}(-1)^{k+1-l}\prod_{\substack{1\leq i\leq k+1\\i \neq l}}(z_{i}-z_{l}-1)\prod_{\substack{1\leq i

That is where I get stuck, I don't find a way to end this induction... the right hand side of what I am trying to prove is the Vandermonde determinant, so there is maybe a miraculous formula about determinant or pfaffian that I could use but I have no clue...

What remains to be proved is the following $$\sum_{l=1}^{k+1}(-1)^{k+1-l}\prod_{\substack{1\leq i\leq k+1\\i \neq l}}(z_{i}-z_{l}-1)\prod_{\substack{1\leq i

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