Integration By Parts of Gamma Function A textbook I'm self-studying - Introduction to Mathematical Statistics by Hogg - has the following text:

T(a) = $\int_{0}^{\infty} y^{\alpha-1}e^{-y} dy$ [gamma function]
If $\alpha > 1$, an integration by parts shows that $T(\alpha) = (\alpha - 1)\int_{0}^{\infty} y^{\alpha-2}e^{-y} dy$.

I'm having trouble deriving this for myself.  How does an Integration by Parts lead to this result?
Thank you.

Let $dv=e^{-x}\ dx$ and $u=x^{\alpha-1}$.  It thus follows that
$$v=-e^{-x},\quad du=(\alpha-1)x^{\alpha-2}\ dx$$
Thus,
$$\begin{align}\int x^{\alpha-1}e^{-x}\ dx&=\int u\ dv\\&=uv-\int v\ du\\&=-x^{\alpha-1}e^{-x}+(\alpha-1)\int x^{\alpha-2}e^{-x}\ dx\end{align}$$
And as the bounds go to $[0,\infty)$, we find that $uv\to0$, hence
$$\int_0^\infty x^{\alpha-1}e^{-x}\ dx=(\alpha-1)\int_0^\infty x^{\alpha-2}e^{-x}\ dx$$
