Multiplying three factorials with three binomials in polynomial identity I have checked the following identity (1) below for $n\leq 40$ with a computer.
Let $(n)_k$ denote the falling factorial $n(n-1)\ldots (n-k+1)$, let
$Z_n=\sum_{k=0}^n (n)_k x^{n-k}$, and finally let
$D_n=x^2Z_{n-1}Z_{n-2}-2nxZ_nZ_{n-2}+Z_nZ_{n-1}$. Then, I conjecture that
$$
D_n=\sum_{0 \leq i \leq j \leq n-2}
(n-j)!(n-j-1)!(j-i)! 
\binom{n-2}{j}\binom{j}{i}\binom{2n+1-i-j}{j-i}x^{i+j}\tag{1}
$$  
I tried my usual tools (induction, WS method, using similar identities) and failed. Any help appreciated.
This identity appears naturally in trying to answer this other MSE question.

It seems there is a flaw in the stated problem. Maybe you want to check it although some time passed by since it was posted.
The polynomials $Z_n$ with
\begin{align*}
Z_n=\sum_{k=0}^n n^{\underline{k}}x^{n-k}
\end{align*}
start with
\begin{align*}
Z_0&=0^{\underline{0}}x^0=1\\
Z_1&=1^{\underline{0}}x^1+1^{\underline{1}}x^0=x+1\\
Z_2&=2^{\underline{0}}x^2+2^{\underline{1}}x^1+2^{\underline{2}}x^0=x^2+2x+4
\end{align*}

We obtain for $n=2$
  \begin{align*}
x^2&Z_{n-1}Z_{n-2}-2nxZ_nZ_{n-2}+Z_nZ_{n-1}\\
&=x^2Z_1Z_0-4xZ_2Z_0+Z_2Z_1\\
&=x^2(x+1)-4x(x^2+2x+4)+(x^2+2x+4)(x+1)\\
&=-2x^3-4x^2-10x+4\tag{2}
\end{align*}

On the other hand

we obtain in (1) for $n=2$
  \begin{align*}
D_2&=\sum_{j=0}^{n-2}\sum_{i=0}^j(n-j)!(n-j-1)!(j-i)!\binom{n-2}{j}\binom{j}{i}\binom{2n+1-i-j}{j-i}x^{i+j}\\
&=\sum_{j=0}^{0}\sum_{i=0}^j(2-j)!(1-j)!(j-i)!\binom{0}{j}\binom{j}{i}\binom{5-i-j}{j-i}x^{i+j}\\
&=2!1!0!\binom{0}{0}\binom{0}{0}\binom{5}{0}x^0\\
&=2
\end{align*}
  which does not coincide with (2).

