The moment generating function of a gamma random variable is $M_X(t)=(1-t/\beta)^{-\alpha}$, we have $E[X]=\alpha/\beta$
$E[X^2]=\frac{\alpha(\alpha+1)}{\beta^2}$
$E[X^3]=\frac{\alpha(\alpha+1)(\alpha+2)}{\beta^3}$
$E[X^4]=\frac{\alpha(\alpha+1)(\alpha+2)(\alpha+3)}{\beta^4}$
and $M_{\bar{X}}(t)=(1-\frac{1}{n\beta}t)^{-n\alpha}$, gives
$E[\bar{X}]=\alpha/\beta$
$E[\bar{X}^2]=\frac{\alpha(\alpha+\frac{1}{n})}{\beta^2}$
$E[\bar{X}^3]=\frac{\alpha(\alpha+\frac{1}{n})(\alpha+\frac{2}{n})}{\beta^3}$
$E[\bar{X}^4]=\frac{\alpha(\alpha+\frac{1}{n})(\alpha+\frac{2}{n})(\alpha+\frac{3}{n})}{\beta^4}$
By using the above identities, we obtain
$E[S^4]=\frac{\alpha}{\beta^4(n-1)^2}\bigg[(n^2-1)\alpha+6(\sqrt{n}-\frac{1}{\sqrt{n}})^2\bigg]$.
I want to find $var(S^2)$, where $S$ is standard deviation. But i don't how to find $E[S^4]$.