Power Series Simplification $N$ is Poisson with parameter $z$. 
Find $E[N\cdot(N-1)\cdot(N-2)\cdots(N-k+1)]$
The answer is apparently $z^k$. 
$$E[N\cdot(N-1)\cdot(N-2)\cdots(N-k+1)]=E\left[\frac{N!}{k!}\right]=\sum_{n\geq0} \frac{n!}{k!}\frac{e^{-z}z^n}{n!}=\sum_{n\geq 0}\frac{e^{-z}z^n}{k!}$$
Where do I go from here?
 A: This is called the $k$th factorial moment of $N$.
\begin{align}
& E[N(N-1)(N-2)\cdots(N-k+1)]  \\[10pt]
= {} & \sum_{n\ge k} \frac{n!}{(n-k)!}\cdot\frac{e^{-z} z^n}{n!} = z^k e^{-z} \sum_{n\,:\,n\ge k} \frac{z^{n-k}}{(n-k)!} = z^k e^{-z} \sum_{\ell\ge 0} \frac{z^\ell}{\ell!} = z^k.
\end{align}
Notice that:


*

*The product $n(n-1)(n-2)\cdots(n-k+1)$ is equal to the fraction $\dfrac{n!}{(n-k)!}$ if $n\ge k$, but not if $n<k$;

*In the sum $\displaystyle\sum_{n\,:\,n\ge k} \frac{e^{-z} z^k z^{n-k}}{(n-k)!}$, the factors $e^{-z}$ and $z^k$ do not change as $n$ runs through the list of integers $\ge k$.  That is what justifies pulling them out of the sum.


You shouldn't be using $k$ and $K$ interchangeably.  In what you've written, $N$ is a random variable and $n$ is the index that runs through the list of its possible values, and it is correct that they should be represented by different letters, and using the capital for the former and lower case for the latter is conventional.  But no such consideration applies to $K$ and $k$.
A related exercise:
Suppose $M$ is the number of fixed points of a uniformly distributed random permutation of $L$ objects.  Show that the first $L$ moments of $M$ are equal to the first $L$ moments of a Poisson distribution with expected value $1$.
HINT: The first $L$ moments are equal if the first $L$ factorial moments are equal.
