How can I prove this for Poisson distribution?

Question

If $X$ has a $P(n\lambda)$ distribution, Then $E[(X)_r]=(n\lambda)^r$ where $(X)_r = \frac{X!}{(X-r)!}$

Answer 1

Lemma: $(x)_r=0$ when $x<r.$

Explanation and proof: If we define $(x)_r = \dfrac{x!}{(x-r)!}$ even if $x<r,$ then we get factorials of negative integers in the denominator and that is problematic. But we have $$ (x)_r = \frac{x(x-1)(x-2)\cdots(x-r+1)}{r(r-1)(r-2)\cdots1}, $$ and using that as the definition, the lemma follows. And that makes sense in combinatorics: the number of permutations of $r$ things from a set of $x$ is $0$ when $x<r.$

And now the proof of the main proposition: \begin{align} \operatorname E((X)_r) = {} & \sum_{x=0}^\infty (x)_r \Pr(X=x) \\[6pt] = {} & \sum_{x=r}^\infty (x)_r\Pr(X=x) \\ & \text{since } (x)_r = 0 \text{ when } x<r \\[8pt] = {} & \sum_{x=r}^\infty \frac{x!}{(x-r)!} \cdot \frac{(n\lambda)^x e^{-n\lambda}}{x!} \\[6pt] = {} & (n\lambda)^r \sum_{x=r}^\infty \frac{(n\lambda)^{x-r} e^{-n\lambda}}{(x-r)!} \end{align} Now let $y=x-r,$ so that as $x$ goes from $r$ to $\infty,$ $y$ goes from $0$ to $\infty,$ and where you see $x-r$ in the sum, put $y$ in its place. Then you'll see that it adds up to $1.$