My proof:
Let $X\sim \mathrm{Poisson}(\lambda)$.
Then $$f_{\Tiny{X}}(x) = \frac{\lambda^x}{x!} e^{-\lambda}.$$
Thus, $E(X) = \sum_{x=0}^{\infty} x f_{\Tiny{X}}(x) = \sum_{x=0}^{\infty} x \frac{\lambda^x}{x!} e^{-\lambda} = \lambda e^{-\lambda}\sum_{x=0}^{\infty} \frac{\lambda^{x-1}}{(x-1)!} = \lambda e^{-\lambda}e^{\lambda}=\lambda$.
$\hspace{16cm}\blacksquare$
I know that the result of the expected value of $X$ is $\lambda$. However, I don't think that $(x-1)!$ exists (to my knowledge) when $x = 0$.
Then $\sum_{x=0}^{\infty} \frac{\lambda^{x-1}}{(x-1)!}$ doesn't exactly make sence since Poisson random variable $X$ takes value from $\{0, 1, 2,\dots\}$.
I have noticed that some of the posts for proving $E(X)= \lambda$ change the summation from $x=0$ to $x=1$ when they cancel $x$ with $\frac{1}{x!}$.
Can anyone explain to me that why you can just change the initial value for the summation?
Many thanks,
Sebastian