I am trying to calculate $$ \text{E}[X(X - 1) \ldots (X - k + 1)], $$ where $ \text{E} $ denotes the expectation operator and $ k \in \mathbb{N} $ is fixed.
I think I have to use the fact that the expectation of a sum of random variables is the sum of the expectations, but how can we apply it to this product?
For a Poisson distribution
$$E[g(X)] = \sum_{j=0}^{\infty} g(j) \frac{\lambda^j}{j!} e^{-\lambda}$$
Specifically, for the $g$ you specified, the sum begins at $j=k$ because $g(X) = 0$ when $X<k$:
$$\begin{align}E[g(X)] &= \sum_{j=k}^{\infty} \frac{j!}{(j-k)!} \frac{\lambda^j}{j!} e^{-\lambda}\\ &= \sum_{j=0}^{\infty} \frac{\lambda^{j+k}}{j!} e^{-\lambda}\\ &= \lambda^k\\ \end{align}$$
Alternatively, the probability generating function $G$ for a Poisson random variable with mean $\lambda$ is $G(s)=E(s^X)=\exp(-\lambda(1-s))$. Differentiate $k$ times and set $s=1$ to get the answer.
