I know that ${n \choose k}$ $ = $ ${n \choose n-k}$ $ = \frac{n!}{(n-k)!(k)!}$, so I would write $$ \sum_{k=0}^{n} {n \choose k} \cdot {n \choose k} = \sum_{k=0}^{n} \frac{n!}{(n-k)!(k)!} \cdot \frac{n!}{(n-k)!(k)!} = \sum_{k=0}^{n} \frac{(n!)^2}{((n-k)!(k)!)^2}$$ however from this point I do not know how to get it in terms of ${n \choose k}$. My book claims that this is equal to ${2n \choose n}$, but I can not see the connection between the two. Even when I write out ${2n \choose n}$, I do not see anything that is connection these two. $${2n \choose n} = \frac{2n!}{(2n-n)!(n)!} = \frac{2n!}{(n)!(n)!} = \frac{2n!}{((n)!)^2}$$
edit: Indeed the book claims indeed that $\sum_{k=0}^{n} {n \choose k} \cdot {n \choose k} = {2n \choose n}$ what has been suggested in the comments
