Integration by parts for evaluation of $\frac{F(m)}{F(m-1)}$

Question

Let $$F(m) = \int_{0}^{\infty} x^{m-1}e^{-x} \,\mathrm{d}x.$$ By integration by parts, evaluate and simplify $\dfrac{F(m)}{F(m-1)}$.

I've tried evaluating $\dfrac{F(m)}{F(m-1)}$ simply by plugging $m$ and $m-1$ into the expressions, and using $$\int uv dx = u\int v dx - \int\int v \frac{du}{dx} dx,$$ which does not really work out.

Answer 1

Hint $$\int_0^{\infty} x^{m-1}e^{-x}dx = -x^{m-1} e^{-x} \big ]_0^{\infty} + \int_0^{\infty} (m-1)x^{m-2} e^{-x}dx $$

Answer 2

Integration by parts is the way to go, we have \begin{align*} F(m)&=\int_0^\infty x^{m-1}e^{-x}\,\mathrm{d}x\\ &=-x^{m-1}e^{-x}\bigg\rvert_{0}^{\infty}+\int_{0}^{\infty}(m-1)x^{m-2}e^{-x}\,\mathrm{d}x\\ &=(m-1)\int_{0}^{\infty}x^{m-2}e^{-x}\,\mathrm{d}x\\ &=(m-1)F(m-1). \end{align*} From this we have $$\frac{F(m)}{F(m-1)}=\frac{(m-1)F(m-1)}{F(m-1)}=m-1$$