I think that follows from a clever transformation of the Beta function I saw some days ago around here:
Let
$B(x,y)=\int_0^1 t^{x-1} (1-t)^{y-1} dt$
Set $t = \dfrac{1}{\tau +1}$, so that
$\eqalign{ & B(x,y) = - \int_\infty ^0 {{{\left( {\frac{1}{{\tau + 1}}} \right)}^{x - 1}}} {\left( {1 - \frac{1}{{\tau + 1}}} \right)^{y - 1}}\frac{{d\tau }}{{{{\left( {\tau + 1} \right)}^2}}} \cr & B(x,y) = \int_0^\infty {{{\left( {\frac{1}{{\tau + 1}}} \right)}^{x - 1}}} {\left( {\frac{\tau }{{\tau + 1}}} \right)^{y - 1}}\frac{{d\tau }}{{{{\left( {\tau + 1} \right)}^2}}} = \int_0^\infty {\frac{{{\tau ^{y - 1}}}}{{{{\left( {\tau + 1} \right)}^{x + y}}}}d\tau } \cr} $
Similarily, let
$\eqalign{ & t = \frac{\tau }{{\tau + 1}} \cr & dt = \frac{{d\tau }}{{{{\left( {\tau + 1} \right)}^2}}} \cr} $
$\eqalign{ & B(x,y) = \int_0^\infty {{{\left( {\frac{\tau }{{\tau + 1}}} \right)}^{x - 1}}} {\left( {1 - \frac{\tau }{{\tau + 1}}} \right)^{y - 1}}\frac{{d\tau }}{{{{\left( {\tau + 1} \right)}^2}}} \cr & B(x,y) = \int_0^\infty {{{\left( {\frac{\tau }{{\tau + 1}}} \right)}^{x - 1}}} {\left( {\frac{1}{{\tau + 1}}} \right)^{y - 1}}\frac{{d\tau }}{{{{\left( {\tau + 1} \right)}^2}}} = \int_0^\infty {\frac{{{\tau ^{x - 1}}}}{{{{\left( {\tau + 1} \right)}^{x + y}}}}d\tau } \cr} $
Can you use that to show the result?