Derivatives of Moment Generating Function of Gamma Distribution

Theorem
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.

Let $t < \beta$.

Let $M_X$ denote the moment generating function of $X$.

The $n$th derivative of $M_X$ is given by:
 * ${M_X}^{\paren n} = \dfrac {\alpha^{\overline n} \beta^\alpha} {\paren {\beta - t}^{\alpha + n} }$

where $\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.

Proof
The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * ${M_X}^{\paren n} = \dfrac {\alpha^{\overline n} \beta^\alpha} {\paren {\beta - t}^{\alpha + n} }$

Basis for the Induction
$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * ${M_X}^{\paren k} = \dfrac {\alpha^{\overline k} \beta^\alpha} {\paren {\beta - t}^{\alpha + k} }$

from which it is to be shown that:
 * ${M_X}^{\paren {k + 1} } = \dfrac {\alpha^{\overline {k + 1} } \beta^\alpha} {\paren {\beta - t}^{\alpha + k + 1} }$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \Z_{\ge 0}: {M_X}^{\paren n} = \dfrac {\alpha^{\overline n} \beta^\alpha} {\paren {\beta - t}^{\alpha + n} }$