Variance of Poisson Distribution/Proof 2

Theorem
Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.

Then the variance of $X$ is given by:
 * $\operatorname {var} \left({X}\right) = \lambda$

Proof 2
From Variance of Discrete Random Variable from PGF, we have:
 * $\operatorname {var} \left({X}\right) = \Pi''_X \left({1}\right) + \mu - \mu^2$

where $\mu = E \left({x}\right)$ is the expectation of $X$.

From the Probability Generating Function of Poisson Distribution, we have:
 * $\Pi_X \left({s}\right) = e^{-\lambda \left({1-s}\right)}$

From Expectation of Poisson Distribution, we have:
 * $\mu = \lambda$

From Derivatives of PGF of Poisson Distribution, we have:
 * $\Pi''_X \left({s}\right) = \lambda^2 e^{- \lambda \left({1-s}\right)}$

Putting $s = 1$ using the formula $\Pi''_X \left({1}\right) + \mu - \mu^2$:
 * $\operatorname {var} \left({X}\right) = \lambda^2 e^{- \lambda \left({1-1}\right)} + \lambda - \lambda^2$

and hence the result.