Consider the integral

∫0∞e−st dt=1se−st∣0∞=1s. \int_0^\infty e^{- st} \ dt = \frac{1}{s} e^{- st} \mid_0^\infty = \frac{1}{s}.

Taking derivative with respect to ss of both side, we get

∫0∞te−ts dt=1s2. \int_0^\infty t e^{- t s} \ dt = \frac{1}{s^2}.

Repeating this process one more step,

∫0∞t2e−st dt=2s3. \int_0^\infty t^2 e^{- st} \ dt = \frac{2}{s^3}.

By induction on k∈N+k \in \mathbb{N}^+, we obtain

∫0∞tke−st dt=k!sk+1. \int_0^\infty t^k e^{- st} \ dt = \frac{k!}{s^{k + 1} }.

Setting s=1s = 1, we have

k!=∫0∞tke−t dt. k! = \int_0^\infty t^k e^{ -t } \ dt.

□\square