Here's a proof of $\mathrm{Var}(X) = \lambda$ for the Poisson distribution.
Proof: First we work out $E(X(X-1))$
$E(X(X-1)) = \sum_Xx(x-1)f(x)$
$= \sum_{x = 0}^{\infty}x(x-1)e^{-\lambda}\frac{\lambda^x}{x!}$
Start from $x = 2$
$= \lambda^2 e^{-\lambda} \sum_{x = 2}^{\infty}\frac{\lambda^{x-2}}{(x-2)!}$
Letting $y = x - 2$ we have
$= \lambda^2 e^{-\lambda} \sum_{y = 0}^{\infty}\frac{\lambda^y}{y!}$
$= \lambda^2 e^{-\lambda} e^\lambda$
$= \lambda^2$
Then
$\mathrm{Var}(X) = EX^2 - (EX)^2$
$= E(X(X - 1) + EX - (EX)^2$
$=\lambda^2 + \lambda - \lambda^2$
$=\lambda$
Question
It's the line after "Start from $x = 2$" that I have an issue with. As far as I can see we have to start from $x = 2$ because otherwise we will have a negative factorial in the denominator which is undefined. But shouldn't that mean the equation itself is undefined, rather than just 'dropping' the undefined variables in the summation and continuing on as we do in the above proof? Why are we allowed to drop the undefined variables in the summation?