Ever wonder how the Euler's Factorial Integral \[\begin{aligned} \int_0^\infty x^n e^{-x} dx = n! \end{aligned}\] can be obtained from \(\displaystyle \int_0^\infty e^{-x} dx \) ?

A great practice of repeated integration by parts: \[\begin{aligned} \int_0^\infty e^{-x} dx = -e^{-x}\big\vert_0^\infty &= 1 \\ \int_0^\infty e^{-x} dx = \int_0^\infty e^{-x} x' dx = x e^{-x}\big\vert_0^\infty + \int_0^\infty x e^{-x} dx &= \int_0^\infty x e^{-x} dx \\ \int_0^\infty (\frac{x^2}{2})' e^{-x} dx = \frac{x^2}{2} e^{-x}\big\vert_0^\infty + \frac{1}{2} \int_0^\infty x^2 e^{-x} dx &= \frac{1}{2} \int_0^\infty x^2 e^{-x} dx \\ \frac{1}{2} \int_0^\infty (\frac{x^3}{3})' e^{-x} dx = \frac{1}{2}\cdot \frac{x^3}{3} e^{-x}\big\vert_0^\infty + \frac{1}{2\cdot 3} \int_0^\infty x^3 e^{-x} dx &= \frac{1}{3!} \int_0^\infty x^3 e^{-x} dx \\ &\vdots \\ \frac{1}{(n-1)!} \int_0^\infty (\frac{x^n}{n})' e^{-x} dx = \frac{1}{(n-1)!} \frac{x^n}{n} e^{-x}\big\vert_0^\infty + \frac{1}{(n-1)!} \frac{1}{n} \int_0^\infty x^n e^{-x} dx &= \frac{1}{n!} \int_0^\infty x^n e^{-x} dx \\ \therefore \frac{1}{n!} \int_0^\infty x^n e^{-x} dx &= 1 \\ \int_0^\infty x^n e^{-x} dx &= n! \end{aligned}\] \(\Box\)

But wait. We can do better than this!

See Differentiating under the integral.