Evaluate the integral $\int^{\infty}_{0} e^{-x}x^{100}dx$

Use a trick which, if I remember correctly, is due to Feynman:

For $\alpha > 0 $, let $$I(\alpha) = \int_{0}^{\infty} e^{-\alpha x} dx = \frac{1}{\alpha}$$

Differentiating under the integral sign $n$ times (with respect to $\alpha$) gives $$\int_{0}^{\infty} (-1)^nx^ne^{-\alpha x}dx = (-1)^n\frac{n!}{\alpha^n} $$ whence $$\int_{0}^{\infty} x^ne^{-\alpha x}dx = \frac{n!}{\alpha^n}. $$ Taking $n=100$ and $\alpha = 1$, shows that the sought integral is equal to $100!$.

Hint Consider the integral $$I(n) := \int_0^{\infty} x^n e^{-x} dx$$ (NB with a general exponent $n$ of $x$). Applying integration by parts with $u = x^n$, $dv = e^{-x} dx$, gives $$\int_0^{\infty} x^n e^{-x} dx = \left. x^n e^{-x} \right\vert_0^{\infty} + n \int_0^{\infty} x^{n - 1} e^x .$$ We can show that the first term is zero (for example, with an inductive argument and L'Hopital's Rule), and so $$\int_0^{\infty} x^n e^{-x} dx = n \int_0^{\infty} x^{n - 1} e^x ,$$ which we can write suggestively as the recurrence relation $$I(n) = n I(n - 1).$$

So, we can evaluate our original integral, which by definition is $I(100)$, by writing $$I(100) = 100 \cdot I(99) = 100 \cdot 99 \cdot I(98) = \cdots.$$ On the other hand, it's easy to compute that $I(0) = 1$.

Notice that $\Gamma(t) = \int\limits_0^\infty \mathrm{e}^{-x}x^{t-1} \,\mathrm{d}x$ is the gamma function. Therefore, $$ \int\limits_0^\infty \mathrm{e}^{-x}x^{100} \,\mathrm{d}x = \Gamma(101) = 100! $$