Let $n \in \mathbb{N}$. Using the factorization $$( X + 1 ) ^ { 2 n } = ( X + 1 ) ^ { n } ( X + 1 ) ^ { n }$$ show that $$\sum _ { k = 0 } ^ { n } \binom { n } { k } ^ { 2 } = \binom { 2 n } { n }$$
Let $n \in \mathbb{N}$. Using the factorization
$$( X + 1 ) ^ { 2 n } = ( X + 1 ) ^ { n } ( X + 1 ) ^ { n }$$
show that
$$\sum _ { k = 0 } ^ { n } \binom { n } { k } ^ { 2 } = \binom { 2 n } { n }$$