Integrating $\int_0^{\infty} u^n e^{-u} du $ I have to work out the integral of
$$
I(n):=\int_0^{\infty} u^n e^{-u} du
$$
Somehow, the answer goes to 
$$
I(n) = nI(n - 1)$$
and then using the Gamma function, this gives $I(n) = n!$
What I do is this:
$$
I(n) = \int_0^{\infty} u^n e^{-u} du
$$
Integrating by parts gives
$$
I(n) =  -u^ne^{-u} + n \int u^{n - 1}e^{-u} $$
Clearly the stuff in the last bit of the integral is now $I(n - 1)$, but I don't see how using the limits gives you the answer. I get this
$$
I(n) = \left( \frac{-(\infty)^n}{e^{\infty}} + nI(n - 1) \right) - \left( \frac{-(0)^n}{e^{0}} + nI(n - 1) \right)
$$
As exponential is "better" than powers, or whatever its called, I get
$$
I(n) = (0 + I(n - 1)) + ( 0 + nI(n - 1)) = 2nI(n - 1)$$
Does the constant just not matter in this case?
Also, I do I use the Gamma function from here? How do I notice that it comes into play? Nothing tells me that $\Gamma(n) = (n - 1)!$, or does it?

You have
$$
I(n) = \lim_{u\to +\infty}u^ne^{-u}-0^ne^{-0}+nI(n-1)
$$
But $$\lim_{u\to +\infty}u^ne^{-u}=\lim_{u\to +\infty}\frac{u^n}{e^{u}}=...=0$$
and so
$$
I(n) =0-0+nI(n-1)=nI(n-1)
$$
