Expectation of Poisson Distribution/Proof 2

Proof
From Probability Generating Function of Poisson Distribution:
 * $\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$

From Expectation of Discrete Random Variable from PGF:
 * $\expect X = \map {\Pi'_X} 1$

We have:

Plugging in $s = 1$:


 * $\map {\Pi'_X} 1 = \lambda e^{- \lambda \paren {1 - 1} } = \lambda e^0$

Hence the result from Exponential of Zero:
 * $e^0 = 1$