$E(X!)$ for Poisson distribution

Question

For $X\sim\text{Pois}(λ)$, find $E(X!)$ (the average factorial of $X$), if it is finite.

Solution:

By LOTUS,

$$E(X!) =e^{−λ}\sum_{k=0}^{\infty} k!\frac{\lambda^k}{k!} = \frac{e^{−\lambda}}{1−\lambda}$$

for $0<\lambda<1$.

Question: Why is just the $k$ on the bottom of the fraction factorial, and not $\lambda^{k!}$ ?

Answer 1

$EX!=\sum P\{X=k\} k!$ and $P\{X=k\}=e^{-\lambda} \frac {\lambda^{k}} {k!}$.

Answer 2

Here's the sum:

$${\cal E}[n!] = \sum\limits_{n=0}^\infty {e^{-\mu} \mu^n \over n!} n! = {e^{-\mu} \over 1 - \mu}$$