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> Let $X$ be an exponentially distributed random variable with parameter $\beta$. Compute $E(X^8)$. Show your work.
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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?