I am trying to show that the expectation of a Poisson variable is equal to the expected value $\lambda$. This is my proof: $$\mathbb{E}[\mathrm{X}]=\sum_{k=0}^\infty k\:\frac{\lambda^k}{k!}e^{-\lambda}=\sum_{k=1}^\infty\frac{\lambda^k}{(k-1)!}e^{-\lambda};$$ now I let $n=k-1$: $$\sum_{n+1}^\infty\frac{\lambda^{n+1}}{n!}e^{-\lambda}=\sum_{n+1}^\infty\lambda\frac{\lambda^{n}}{n!}e^{-\lambda}=\lambda e^{\lambda}e^{-\lambda}=\lambda.$$
This seems right to me, but there is one thing that bothers me: In the last sum where I apply the Taylor series definition of the exponential function, the index of the summation begins at $n+1$, rather than $n$. This makes me doubt whether I can apply the definition to get the proof. Thank you.
If $n = k-1$, then $$\sum_{k=1}^\infty \frac{\lambda^k}{(k-1)!} e^{-\lambda} = \sum_{\color{red}{n = 0}}^\infty \frac{\lambda^{n+1}}{n!} e^{-\lambda}.$$ It doesn't really make sense to just write $n+1$ below the summation sign.
Think of it this way: if the summation starts at $k = 1$, and $n = k-1$, then $n$ starts at $0$. The terms look like this:
$$\frac{\lambda^1}{(1-1)!} e^{-\lambda} + \frac{\lambda^2}{(2-1)!} e^{-\lambda} + \cdots$$ which of course is the same as saying $$\frac{\lambda^0}{0!} \lambda e^{-\lambda} + \frac{\lambda^1}{1!} \lambda e^{-\lambda} + \cdots$$