Show that : $$\beta \left( x,n\right) =\dfrac {\left( n-1\right) !}{x\left( x+1\right) \left( x+2\right) ....\left( x+n-1\right) }$$
My attempt : $$\beta \left( x,n\right) =\dfrac {\Gamma \left( x\right) \Gamma \left( n\right) }{\Gamma \left( x+n\right) }$$
$$\Gamma \left( n\right) =\left( n-1\right) !$$
We have
$$\beta(x,n)=\int_0^1 t^{x-1}(1-t)^{n-1}dt$$ so by an integration by parts we find $$\beta(x,n)=(n-1)\frac1x\int_0^1 t^x(1-t)^{n-2}dt=\frac{n-1}x \beta(x+1,n-1)$$ Now a simple induction gives the desired result.
Remark In the induction you reach to: $$\beta(x+n-1,1)=\int_0^1 t^{x+n-2}dt=\frac{1}{x+n-1}$$