Trying to understand a result in a representation theoretical paper, I realized that it implies the following elementary identity for symmetric functions. **My question is whether this identity is true, known, and whether it has a direct proof.** It might also happen that this identity is simpler than I think.


Let $d$ be a natural number. For $0\leq k\leq d$ let $e\_k(z\_1\dots,z\_d)$ denote the $k$th elementary symmetric polynomial in $d$ variables, namely
$$e\_k(z\_1\dots,z\_d)=\sum\_{1\leq i\_1<\dots <i\_k\leq d}z\_{i\_1}\dots z\_{i\_k}.$$


Let $x$ be a variable. **The relevant identity is
\begin{eqnarray\*}
\sum\_{k=0}^d (-1)^k(x-d)^{2k}\cdot e\_{d-k}((x-d+1)^2,\dots,(x-1)^2,x^2)=\\
d!\cdot (2x-2d+1)\dots (2x-d-1)
(2x-d).\end{eqnarray\*}**


Notice that the right hand side is a polynomial in $x$ of degree $d$, while the left hand side has a priori degree at most $2d$ only.



An identity for elementary symmetric functions
Viete formula gives
$\sum\_{k=0}^{d} t^k e\_{d-k}\left(\alpha\_0,\ldots,\alpha\_{d-1}\right)=(t+\alpha\_1)\ldots(t+\alpha\_d).$


Therefore (substituing $t=-(x-d)^2$ and $\alpha\_j=(x-j)^2$), LHS is equal to 
$\prod\_{j=0}^{d-1} \left(-(x-d)^2+(x-j)^2\right)=
\prod\_{j=0}^{d-1} (2x-j-d)(d-j)=d!\prod\_{k=d}^{2d-1} (2x-k)$


which is exactly the RHS.


