Differentiating under the integral sign

1. $I(a)=\left[-\frac1ae^{-ax}\right]_0^{+\infty}=\frac1a$.

2. Differentiating $I(a)=\int_0^\infty e^{-ax}dx$ in $a$ gives $I'(a)=-\int_0^\infty xe^{-ax}dx$, while $I'(a)=-\frac1{a^2}$. Hence
$$\int_0^{+\infty}xe^{-ax}\,dx=\frac1{a^2}.$$
Repeating $n$ times, $(-1)^nI^{(n)}(a)=\int_0^\infty x^ne^{-ax}dx$ and $I^{(n)}(a)=(-1)^n\frac{n!}{a^{n+1}}$, so
$$\int_0^{+\infty}x^ne^{-ax}\,dx=\frac{n!}{a^{n+1}}.$$

3. Let $G_n=\int_0^\infty x^ne^{-x}dx$. By parts with $u=x^n$, $v'=e^{-x}$:
$$G_n=\bigl[-x^ne^{-x}\bigr]_0^{+\infty}+n\int_0^\infty x^{n-1}e^{-x}\,dx=nG_{n-1},$$
since $x^ne^{-x}\to0$ at infinity. With $G_0=1$, $G_n=n!$.

Remark. The function $\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}dx$ extends the factorial to real numbers: $\Gamma(n+1)=n!$.