Show that for X ~ Gamma($\alpha$, $\beta$), for positive constant $\nu$,
$E[X^\nu] = \dfrac{\beta^\nu*\Gamma(\nu + \alpha)}{\Gamma(\alpha)}$.
I have the following solution:
However, I don't understand how we get that $\dfrac{1}{\Gamma(\alpha)*\beta^\alpha}*\int_{0}^{\infty}x^{(\nu+\alpha)-1}e^{-x/\beta}dx = \dfrac{\Gamma(\nu+\alpha)*\beta^{\nu+\alpha}}{\Gamma(\alpha)*\beta^{\alpha}}$
Would appreciate any help on how this step was completed. Basically, I don't understand how: $\int_{0}^{\infty}x^{(\nu+\alpha)-1}e^{-x/\beta}dx = \Gamma(\nu+\alpha)*\beta^{\nu+\alpha}$.
I see that the left side is close to the definition of the Gamma function, but can't see how exactly to turn it into the right side.