Suppose I have a random variable $X \sim \text{Pois}(\lambda)$. I want to calculate $E[X^{n}]$, in terms of $n$. Here's what I did:
$$E[x^{n}] = \sum_{x = 0}^{\infty} x^{n} \cdot \frac{\lambda^{x} e^{-\lambda}}{x!}$$
$$= \sum_{x = 0}^{\infty} x \cdot x^{n - 1} \cdot \frac{\lambda \cdot \lambda^{x-1}e^{-\lambda}}{x!} $$
$$= \lambda \sum_{x = 0}^{\infty} x^{n - 1} \cdot \frac{\lambda^{x - 1} e^{-\lambda}}{(x - 1)!}$$
$$= \lambda \sum_{y = -1}^{\infty} (y + 1) \frac{e^{-\lambda}\cdot \lambda^{y}}{y!}$$
$$= \lambda E(X + 1)^{n - 1}.$$
Above, I set $y = x - 1$. Is this correct? Are the bounds on my sum correct? or should it go from $n = 0$ to $\infty$? I know that the answer is right, but i want to check my work, please.
EDIT: Attempt #2
Ok, so the sum should be from $0$ to $\infty$. Is this correct?
$$\mathbb{E}[X^{n}] = \sum_{x = 0}^{\infty} x^{n} \cdot \frac{e^{-\lambda} \lambda^{x}}{x!} $$
$$= \lambda \sum_{x = 1}^{\infty} x^{n - 1} \cdot \frac{e^{-\lambda} \lambda^{x - 1}}{(x - 1)!}$$
I could change the sum from $x = 0 $ to $x = 1$ because the term becomes $0$ for the $x = 0$ index. Now let $y = x - 1$:
$$\lambda \cdot \sum_{y = 0}^{\infty}(y + 1)^{n - 1} \cdot \frac{e^{-\lambda} \lambda^{y}}{y!} $$
$$ = \lambda \cdot E[(X + 1)^{n - 1}].$$
Moments of Poisson distribution can be obtained using this recursive formula:
$$m_n (\lambda) = \mathbb{E}\{x^n\} = e^{-\lambda}\sum_{k=0}^{\infty}k^n \frac{\lambda^k}{k!} \Rightarrow \frac{\partial \; m_n(\lambda)}{\partial \lambda} = -e^{\lambda}\sum_{k=0}^{\infty}k^n \frac{\lambda^k}{k!} + e^{-\lambda}\sum_{k=0}^{\infty}k^(n+1) \frac{\lambda^{(k-1)}}{k!} = -m_n(\lambda)+\frac{1}{\lambda}m_{n+1}(\lambda) \Rightarrow m_{n+1}(\lambda) = \lambda\times\left( m_n(\lambda) + m^{'}_n(\lambda) \right)$$