Theorem: Excess Kurtosis of Gamma Distribution

Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac 6 \alpha$

Proof:
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Gamma Distribution, we have: 
:$\mu = \dfrac \alpha \beta$
By Variance of Gamma Distribution, we have: 
:$\sigma = \dfrac {\sqrt \alpha} \beta$ 
So: 
{{begin-eqn}}
{{eqn	| l = \gamma_2
	| r = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3
	| c = Kurtosis in terms of Non-Central Moments
}}
{{eqn	| r = \frac {\expect {X^4} - 4 \paren {\dfrac \alpha \beta} \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} } + 6 \paren {\dfrac \alpha \beta}^2 \paren {\dfrac {\alpha \paren {\alpha + 1} } {\beta^2} } - 3 \paren {\dfrac \alpha \beta}^4} {\paren {\dfrac {\alpha^2} {\beta^4} } } - 3
	| c = Skewness of Gamma Distribution, Variance of Gamma Distribution
}}
{{end-eqn}}
To calculate $\gamma_2$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$. 
By Moment Generating Function of Gamma Distribution, this is given by: 
:$\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha} = \beta^\alpha \paren {\beta - t}^{-\alpha}$
From Moment in terms of Moment Generating Function:
:$\expect {X^4} = \map {M_X' ' ' '} 0$
So:
{{begin-eqn}}
{{eqn	| l = \map {M_X'} t
	| r = \beta^\alpha \paren {-\alpha } \paren {\beta - t}^{-\alpha - 1} \paren {-1}
	| c = Chain Rule for Derivatives, Derivative of Power
}}
{{eqn	| r = \beta^\alpha \alpha  \paren {\beta - t}^{-\alpha - 1}
	| c = 
}}
{{eqn	| l = \map {M_X' '} t
	| r = \beta^\alpha \alpha \paren {-\alpha - 1}  \paren {\beta - t}^{-\alpha - 2} \paren {-1}
	| c = Chain Rule for Derivatives, Derivative of Power
}}
{{eqn	| r = \beta^\alpha \alpha \paren {\alpha + 1}  \paren {\beta - t}^{-\alpha - 2}
	| c = 
}}
{{eqn	| l = \map {M_X' ' '} t
	| r = \paren {-\alpha - 2} \beta^\alpha \alpha \paren {\alpha + 1}  \paren {\beta - t}^{-\alpha - 3} \paren {-1}
	| c = Chain Rule for Derivatives, Derivative of Power
}}
{{eqn	| r = \beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2}  \paren {\beta - t}^{-\alpha - 3}
	| c = 
}}
{{eqn	| l = \map {M_X' ' ' '} t
	| r = \paren {-\alpha - 3} \alpha \paren {\alpha + 1} \paren {\alpha + 2} \beta^\alpha  \paren {\beta - t}^{-\alpha - 4} \paren {-1}
	| c = Chain Rule for Derivatives, Derivative of Power
}}
{{eqn	| r = \alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} \beta^\alpha  \paren {\beta - t}^{-\alpha - 4}
	| c = 
}}
{{end-eqn}}
Setting $t = 0$:
{{begin-eqn}}
{{eqn	| l = \expect {X^4}
	| r = \alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} \beta^\alpha  \paren {\beta - 0}^{-\alpha - 4}
}}
{{eqn	| r = \dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} } {\beta^4}
	| c = 
}}
{{end-eqn}}
Plugging this result back into our equation above:
{{begin-eqn}}
{{eqn	| l = \gamma_2
	| r = \frac {\expect {X^4} - 4 \paren {\dfrac \alpha \beta} \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} } + 6 \paren {\dfrac \alpha \beta}^2 \paren {\dfrac {\alpha \paren {\alpha + 1} } {\beta^2} } - 3 \paren {\dfrac \alpha \beta}^4} {\paren {\dfrac {\alpha^2} {\beta^4} } } - 3
	| c = 
}}
{{eqn	| r = \frac {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} } {\beta^4} - 4 \paren {\dfrac \alpha \beta} \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} } + 6 \paren {\dfrac \alpha \beta}^2 \paren {\dfrac {\alpha \paren {\alpha + 1} } {\beta^2} } - 3 \paren {\dfrac \alpha \beta}^4} {\paren {\dfrac {\alpha^2} {\beta^4} } } - 3
	| c = 
}}
{{eqn	| r = \frac {\paren {\alpha^4 + 6 \alpha^3 + 11 \alpha^2 + 6 \alpha} - 4 \paren {\alpha^4 + 3 \alpha^3 + 2 \alpha^2}  + 6 \paren {\alpha^4 + \alpha^3 } - 3 \alpha^4} {\paren {\alpha^2 } } - 3
        | c = $\beta^4$ cancels
}}
{{eqn	| r = \frac {\paren {1 - 4 + 6 - 3} \alpha^4 + \paren {6 - 12 + 6} \alpha^3 + \paren {11 - 8} \alpha^2 + 6 \alpha } {\paren {\alpha^2 } } - 3
        | c = 
}}
{{eqn	| r = \dfrac 6 \alpha
}}
{{end-eqn}}
{{qed}}