Variance of Geometric Distribution/Proof 2

Proof
From Variance of Discrete Random Variable from PGF:
 * $\var X = \map {\Pi''_X} 1 + \mu - \mu^2$

where $\mu = \expect X$ is the expectation of $X$.

From the Probability Generating Function of Geometric Distribution:
 * $\map {\Pi_X} s = \dfrac q {1 - p s}$

where $q = 1 - p$.

From Expectation of Geometric Distribution:
 * $\mu = \dfrac p q$

From Derivatives of PGF of Geometric Distribution:
 * $\map {\Pi''_X} s = \dfrac {2 q p^2} {\left({1 - ps}\right)^3}$

Putting $s = 1$ using the formula $\map {\Pi''_X} 1 + \mu - \mu^2$:
 * $\var X = \dfrac {2 q p^2} {\paren {1 - p}^3} + \dfrac p q - \paren {\dfrac p q}^2$

and hence the result, after some algebra.