Calculating $E[Z(Z-1)(Z-2)(Z-3)]$ where $Z$ is Poisson If $Z$ is a variable that distributes by Poisson, with Expected value, $E(Z) = 2.5$. 
I need to solve:
$E[Z(Z-1)(Z-2)(Z-3)]$
So What I thought to do is first:
$E[(Z^2-Z)(Z^2-2Z)(Z^2-3Z)]$
$[E(Z^2)-E(Z)][E(Z^2)-2E(Z)][E(Z^2)-3E(Z)]$
From here I'm not sure how can I insert the $E(Z) = 2.5$ if what I get in the end is $E(Z^2)$

Here is a plain solution, because the exercise was rather designed to have this solution path. We have:
$$
\begin{aligned}
&\Bbb E[\ Z(Z-1)(Z-2)(Z-3)\ ]\\
&=
\sum_{k\ge 0}k(k-1)(k-2)(k-3)\cdot e^{-\lambda}\frac {\lambda^k}{k!}\\
&=
\sum_{k\ge 4}k(k-1)(k-2)(k-3)\cdot e^{-\lambda}\frac {\lambda^k}{k!}\\
&=
\lambda^4\sum_{k-4\ge 0} e^{-\lambda}\frac {\lambda^{k-4}}{(k-4)!}\\
&=\lambda^4\cdot 1\ .
\end{aligned}
$$
With the same argument, the general expectation of a random variable of the shape $Z(Z-1)\dots(Z-(m-1))$ is $\lambda^m$.
