Using the Stein-Chen identity for the Poisson distribution to show that $E[X(X-1)...(X-k+1)] = \lambda^k$ So far, I have already derived the following:
$E[(X-\lambda)g(X)] = \lambda E[g(X+1)-g(X)]$
But I am stuck in terms of applying this identity to derive $\lambda^k$:
\begin{align}
\lambda^k &= E[X(X-1)...(X-k+1)]\\
&= E[X(X-1)]...E[X-k+1]\\
&= 1\cdot E[(X+1)-X]...E[X-k+1] \\
&= 1\cdot E[1]...E[X-k+1]
\end{align}
I know I am doing something wrong here (i.e. $E[X(X-1)] \neq E[1]$) but this is the best I can come up with right now. I guess I am confused on what my g(X) and $\lambda$ are supposed to be. 
 A: We will proceed my induction. For $k=1$ we have $\mathbb E[X] = \lambda$, which is true. Assume that for some $k \in \mathbb N$ we have $\mathbb E[X(X-1)...(X-k+1)] = \lambda^k$.
Stein-Chen identity, tells us that for poisson r.v we have $\mathbb E[Xg(X)] = \lambda \mathbb E[g(X+1)]$.
Take $g:\mathbb R \to \mathbb R$, $g(x) = (x-1)...(x-k)$
Then $\mathbb E[Xg(X)] = \mathbb E[X(X-1)...(X-k)]$, while $\lambda \mathbb E[g(X+1)] = \lambda \mathbb E[X(X-1)...(X-k+1)]$, which is equal to $\lambda \cdot \lambda^k$ by our induction assumption. So, we get $\mathbb E[X(X-1)...(X-k)] = \lambda^{k+1}$
Q.E.D
A: Must you use Stein-Chen? We can derive this result using generating functions. Let $G(s) = \mathbb E[s^X]$, then
\begin{align}
g(S) &= \mathbb E[s^X]\\
&= \sum_{n=0}^\infty s^ne^{-\lambda}\frac{\lambda^n}{n!}\\
&=e^{-\lambda}\sum_{n=0}^\infty \frac{(\lambda s)^n}{n!}\\
&=e^{-\lambda}e^{\lambda s}\\
&=e^{\lambda(s-1)}.
\end{align}
Now, the $k^{\mathrm{th}}$ factorial moment of $X$ is given by
\begin{align}
\mathbb E\left[\frac{X!}{(X-k)!} \right] &= \lim_{s\uparrow1}G^{(k)}(s).\\
&= \lim_{s\uparrow1} \frac{\mathsf d}{\mathsf ds} e^{\lambda(s-1)}\\
&= \lim_{s\uparrow1} \lambda^n e^{\lambda(s-1)}\\
&= \lambda^n.
\end{align}
