There is no general tool for recurrance relations like that.

In your case we have:
F(n)=product from {i=1} to n of (2i-1)(i)
F(n)=1/2^n product from {i=1} to n of (2i-1)(2i)
F(n)=1/2^n product from {i=1} to 2n of (i)
F(n)=1/2^n (2n)!

That said, I haven't the foggiest if I would have been able to do that without seeing your 'closed form' version. Notably, n! is generally not really a closed form sort of thing unless you're using something like stirling's approximation.