A discrete operator begets even/odd polynomials Given a function $f(x)$ define the forward shift operator by $Ef(x)=f(x+1)$ and the discrete derivative $\delta f(x)=(E-1)f(x)=f(x+1)-f(x)$. 
Given a partition $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_k)$, where $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k\geq1$ and $k>0$, define the operator 
$$L_{\lambda}=\frac{(E^{\lambda_1}-1)\cdots(E^{\lambda_k}-1)}{\delta}.$$
Let $(x)_n=x(x-1)\cdots(x-n+1)$ be the falling factorial.

Question.  If $\lambda\vdash n$ then is it true $\Phi_n(x)=L_{\lambda}(x)_n$ is either an even or an odd polynomial, with non-negative integer coefficients? It appears to be so.

Example. If $\lambda=(n)$ then $L_{\lambda}(x)_n=\frac{(x+n)_{n+1}-(x)_{n+1}}{n+1}$ indeed satisfies the claim (check!).
 A: I have found a proof. I hope someone else can give a more conceptual argument.
Let $\Psi_{\lambda}(x)=L_{\lambda}(x)_n$. We approach the expansion of $\Psi$ differently. Begin with
\begin{align}
\prod_{i=1}^k(E^{\lambda_i}-1)
&=\sum_{T\subset\lambda}(-1)^{k-\#T}E^{\vert T\vert} \\
&=\frac12\left(\sum_{T\subset\lambda}(-1)^{k-\#T}E^{\vert T\vert}+
\sum_{T^c\subset\lambda}(-1)^{k-\#T^c}E^{\vert T^c\vert}\right)\\
&=\frac12\sum_{T\subset\lambda}(-1)^{\#T}\left(
(-1)^kE^{\vert T\vert}+E^{n-\vert T\vert}\right);\end{align}
where $\#T=$ the cardinality of $T$ (if empty then zero), $\vert T\vert=$ sum 
of elements of $T$ and $T^c$ is the complement of $T$ in the set $\lambda$.  
The next step uses a couple of key facts, namely:
$$(x+n-q)_{n+1}=(-1)^{n+1}(-x+q)_{n+1} \qquad \text{and} \qquad
\frac1{\delta}(x)_n=\frac{(x)_{n+1}}{n+1}.$$ 
We thus compute
\begin{align}
\Psi_{\lambda}(x)&=\frac1{\delta}\prod_{i=1}^k(E^{\lambda_i}-1)(x)_n\\
&=\frac1{2(n+1)}\sum_{T\subset\lambda}(-1)^{\#T}\left((-1)^k
(x+\vert T\vert)_{n+1}+(x+n-\vert T\vert)_{n+1}\right)\\
&=\frac1{2(n+1)}\sum_{T\subset\lambda}(-1)^{\#T}\left((-1)^k
(x+\vert T\vert)_{n+1}+(-1)^{n+1}(-x+\vert T\vert)_{n+1}\right)\\
&=\frac{(-1)^{n+k+1}}{2(n+1)}\sum_{T\subset\lambda}(-1)^{\#T}\left((-1)^{n+1}
(x+\vert T\vert)_{n+1}+(-1)^k(-x+\vert T\vert)_{n+1}\right)\\
&=(-1)^{n+k+1}\Psi_{\lambda}(-x).\end{align}
The proof is complete. $\square$
