Binomial identity $\binom{0}{0}\binom{2n}{n}+\binom{2}{1}\binom{2n-2}{n-1}+\binom{4}{2}\binom{2n-4}{n-2}+\cdots+\binom{2n}{n}\binom{0}{0}=4^n.$
Proof of a combinatorial identity: $\sum\limits_{i=0}^n {2i \choose i}{2(n-i)\choose n-i} = 4^n$