How does $\tbinom{4n}{2n}$ relate to $\tbinom{2n}{n}$? I got this question in my mind when I was working on a solution to factorial recurrence and came up with this recurrence relation: $$(2n)!=\binom{2n}{n}(n!)^2$$ which made me wonder: is there also a recurrence relation for $\tbinom{4n}{2n}$ in terms of $\tbinom{2n}{n}$? Please use no factorials greater than $(2n)!$, preferably not greater than $n!$.

I think this is what you are looking for (but this is just back substitution):
$$\binom{4n}{2n} = \frac{(4n)!}{((2n)!)^2} = \frac{(4n)!}{\binom{2n}{n}^2 (n!)^4}.$$
Please update me on whether this is what you're looking for...
