Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 17 / b / Solution

For $n>1$, integration by parts with $u=x^{n-1}$ and $dv=e^{-x}dx$ yieldsThe boundary term vanishes at zero because $n-1>0$, and at infinity because exponential decay dominates every fixed power. The initial improper integral is $I_1={\int }_0^{\infty }{e}^{-x}dx=1$. Iterating the recurrence relation therefore givesThis is the positive-integer case of the Gamma function identity $\Gamma (n)=(n-1)!$.