Expectation property for Poisson distribution

Question

Let X be a random variable having a Poisson distribution with parameter $\lambda$. Prove that, for n = 1; 2,....E[X$^{n}$] = $\lambda$E[(X + 1)$^{n-1}$].

This is what I tried to do:

Answer 1

You need to use induction here. For $n=1$ we have $$ \mathbb E[X^1] = \lambda = \lambda\cdot\mathbb E[(X+1)^0]. $$ Assume now that $\mathbb E[X^n] = \lambda\cdot\mathbb E[(X+1)^{n-1}]$ for some positive integer $n$. Then \begin{align} \mathbb E[X^n] &= \sum_{k=0}^\infty \frac{k^n e^{-\lambda}\lambda^k}{k!}\\ &= \lambda\sum_{k=1}^\infty \frac{k^{n-1}e^{-\lambda}\lambda^{k-1}}{(k-1)!}\\ &= \lambda\sum_{k=0}^\infty \frac{(k+1)^{n-1}e^{-\lambda}\lambda^k}{k!}\\ &= \lambda\cdot\mathbb E[(X+1)^{n-1}]. \end{align}