Derivatives of Probability Generating Function at One

Theorem
Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.

Then the $n$th derivative of $\map {\Pi_X} s$ at $s = 1$ is given by:
 * $\dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$

for $n = 1, 2, \ldots$

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $\dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$

Basis for the Induction
$\map P 1$ is the case:
 * $\dfrac \d {\d s} \map {\Pi_X} 1 = \expect X$

This is demonstrated in Expectation of Discrete Random Variable from PGF.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1})$ is true.

So this is our induction hypothesis:
 * $\dfrac {\d^k} {\d s^k} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - k + 1} }$

Then we need to show:
 * $\dfrac {\d^{k + 1} } {\d s^{k + 1} } \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - k} }$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N_{>0}: \dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$