I need some help on this reduction formulae question.

The question is:

Given that $\displaystyle \

I_n = \int_0^{\frac{\pi }{2}} {\sin ^n } xdx

\ $

Show that

$\displaystyle \

I_{2n} = \frac{{(2n)!\pi }}{{2^{2n + 1} (n!)^2 }}

\$

I know that the reduction formulae is

$\displaystyle \

I_n = \frac{{n - 1}}{n}I_{n - 2}

\$

But I don't know how manipulate it into the above form. Could someone help please?