The question:

Show that $\displaystyle \int_0^1 {x^m(1 - x)^n \ dx = \frac{m!n!}{(m + n + 1)!}$ for all nonnegative integers m and n.

My attempt:

Using integration by parts:

$\displaystyle u = x^m$

$\displaystyle u' = mx^{m - 1}$

$\displaystyle v' = (1 - x)^n$

$\displaystyle v = -\frac{(1 - x)^{n + 1}}{n + 1}$

$\displaystyle I_{m,n} = _0^1[-x^m\frac{(1 - x)^{n + 1}}{n + 1}] + m\int_0^1 {x^{m - 1}\frac{(1 - x)^{n + 1}}{n + 1}}$

= $\displaystyle _0^1[-x^m\frac{(1 - x)^{n + 1}}{n + 1}] + \frac{m}{n + 1}\int_0^1 {x^{m - 1}(1 - x)^{n + 1}}$

= $\displaystyle \frac{m}{n + 1}\int_0^1 {x^{m - 1}(1 - x)^{n + 1}}$

= $\displaystyle \frac{m}{n + 1} I_{m-1, n+1}$

I think that's right to far. I'm not sure what to do from here. Any assistance would be most appreciated!