The falling factorials are $(x)\_n=x(x-1)\cdots(x-n+1)$ with $(x)\_0:=1$. Define the forward shift $E$ and the discrete derivative $\delta=E-1$, respectively, by
$$Ef(x)=f(x+1) \qquad \text{and} \qquad \delta f(x)=f(x+1)-f(x).$$
Let $\lambda=(\lambda\_1,\dots,\lambda\_k)\vdash n$ be a partition of a positive integer $n$. Now, define the operator $\Psi\_{\lambda}$ to act on polynomials according to 
$$\Psi\_{\lambda}=(E-1)^{-1}\prod\_{j=1}^{k}(E^{\lambda\_j}-1)$$
(where $k $ is the length of $\lambda$)
so in particular denote $\Psi\_{\lambda}(x):=\Psi\_{\lambda}((x)\_n)$. 


For example, if $n=9$ and $\lambda=(4,2,2,1)$ then $(E-1)^{-1}(E^4-1)(E^2-1)^2(E-1)=(E^4-1)(E^2-1)^2=E^8-2E^6+2E^2-1$ and hence $\Psi\_{\lambda}(x)=(x+8)\_9-2(x+6)\_9+2(x+2)\_9-(x)\_9$, or
$$\Psi\_{\lambda}(x)=8064x^6+141120x^4+213696x^2.$$
Experiments suggest to ask



> 
> is it true that $\Psi\_{\lambda}(x)$ is always an even or an odd polynomial?
> 
> 
> 



These polynomials are always either even or odd
We first note that $\Psi\_\lambda$ is either symmetric or odd-symmetric in $E$. In other words, if we think of $\Psi\_\lambda$ as a polynomial $P$ applied to the operator $E$ (of degree $n - 1$), then $P(t)=\pm t^{n-1} P(\frac{1}{t})$ (plus if the number of elements $|\lambda|$of the partition is odd, minus if even). This is easy to see from your product expression, as every multiplicand is odd-symmetric. Write $P(t)=\sum a\_i t^i$; symmetry says $a\_{n-1-i}=a\_i$, while odd-symmetry says $a\_{n-1-i}=-a\_i$. Together, we have $a\_{n-1-i}=-(-1)^{|\lambda|} a\_i$.


Simple examination shows that $(x)\_n = (-1)^n (n-1-x)\_n$. Therefore:


$$
\begin{aligned}\Psi\_\lambda(x) &= \sum a\_i E^i (x)\_n = -(-1)^{n+|\lambda|} \sum a\_{n-1-i} E^i (n-1-x)\_n \\
&= -(-1)^{n+|\lambda|} \sum a\_{n-1-i} \left(E^{n-1-i} (y)\_n\right)\_{y=-x} = -(-1)^{n+|\lambda|} \Psi\_\lambda(-x),
\end{aligned}$$ 
where the third equality uses the fact that $E$ goes the "other way" for $-x$. 


So your function is odd iff $n+|\lambda|$ is even, and vice versa.


