Calculating the 8th moment of an exponential random variable 
Let $X$ be an exponentially distributed random variable with parameter $\beta$. Compute $E(X^8)$. Show your work.

I'm preparing for a final exam, and currently trying to figure out if my solution to this question is correct, I would greatly appreciate any input. Below is what I've tried:
Since the the exponential distribution is a special case of the gamma distribution with $\alpha=1$, we can write derive the general case $E(X^{k})$:
$\int_0^{\infty}\frac{x^{k}\cdot x^{\alpha-1} \cdot e^{-x/\beta}}{\Gamma(\alpha) \cdot \beta^{\alpha}} dx$ = $\int_0^{\infty}\frac{x^{k+\alpha-1} \cdot e^{-x/\beta}}{\Gamma(\alpha) \cdot \beta^{\alpha}} dx$
We now want to manipulate the expression inside the integral in order to obtain an area equal to one. We do the following:
$\int_0^{\infty}\frac{x^{k+\alpha-1} \cdot e^{-x/\beta}}{\Gamma(\alpha) \cdot \beta^{\alpha}} dx$ = $\frac{\Gamma(\alpha + k) \cdot \beta^{k}}{\Gamma(\alpha)} \int_0^{\infty}\frac{x^{k+\alpha-1} \cdot e^{-x/\beta}}{\Gamma(\alpha + k) \cdot \beta^{\alpha + k}} dx$
Since the area inside the integral is equal to one, we arrive at: 
$E(X^{k}) = \frac{\Gamma(\alpha + k) \cdot \beta^{k}}{\Gamma(\alpha)}$
With this result, we now try $E(X^8)$:
$E(X^8) = \frac{\Gamma(\alpha + 8) \cdot \beta^{8}}{\Gamma(\alpha)} = \frac{\Gamma(9)\beta^{8}}{(0)!} = (8!) \beta^{8}$.
I am not sure I have performed this last part right. Is there another way to approach this?
 A: Your solution is correct and valid.
Alternate ways of doing this:  note that the MGF of an exponential distribution with scale parameter $\beta$ (as you have defined it in your question) is $$M_X(t) = \operatorname{E}[e^{tX}] = \int_{x=0}^\infty e^{tx} \frac{e^{-x/\beta}}{\beta} \, dx = \frac{1}{\beta} \int_{x=0}^\infty e^{-x(\beta^{-1} - t)} \, dx = \frac{1}{1 - \beta t} \int_{x=0}^\infty \frac{e^{-x(\beta^{-1} - t)}}{(\beta^{-1} - t)^{-1}} \, dx = (1 - \beta t)^{-1},$$ over a domain for $t$ for which the integral is convergent.  Then $$\operatorname{E}[X^k] = \left[ \frac{d^k M_X(t)}{dt^k} \right]_{t=0} = \left[ k! (1-\beta t)^{-k-1} \beta^k \right]_{t=0} = k! \beta^k.$$

Another way:  recognize that $X = \beta Y$ for $Y \sim \operatorname{Exponential}(1)$, so that $$X^k = \beta^k Y,$$ and it is now much easier to perform the integration directly:  $$\operatorname{E}[Y^k] = \int_{y=0}^\infty y^k e^{-y} \, dy \equiv \Gamma(k+1) = k!,$$ as the result follows from the definition of the gamma function.
A: You could instead note that the moment generating function is 
$M(t) = (1 - \beta t)^{-1}$.
Taking derivatives successively, you will see a pattern:
$M'(t) = \beta (1 - \beta t)^{-2}$
$M''(t) = 2 \beta^2 (1 - \beta t)^{-3}$
$\vdots$
$M^(8) (t) = 8! \beta^8 (1 - \beta t)^{-9}$
Evaluating at $t=0$, you recover the value $E(X^8) = 8! \beta^8$.
