Sum of Squares of Binomial Coefficients/Combinatorial Proof

Proof
Consider the number of paths in the integer lattice from $\left({0, 0}\right)$ to $\left({n, n}\right)$ using only single steps of the form:
 * $\left({i, j}\right) \to \left({i + 1, j}\right)$
 * $\left({i, j}\right) \to \left({i, j + 1}\right)$

that is, either to the right or up.

This process takes $2 n$ steps, of which $n$ are steps to the right.

Thus the total number of paths through the graph is equal to $\dbinom {2 n} n$.

Now let us count the paths through the grid by first counting the paths:
 * $(1): \quad$ from $\left({0, 0}\right)$ to $\left({k, n - k}\right)$

and then the paths:
 * $(2): \quad$ from $\left({k, n - k}\right)$ to $\left({n, n}\right)$.

Note that each of these paths is of length $n$.

Since each path is $n$ steps long, every endpoint will be of the form $\left({k, n - k}\right)$ for some $k \in \left\{{1, 2, \ldots, n}\right\}$, representing $k$ steps right and $n-k$ steps up.

Note that the number of paths through $\left({k, n - k}\right)$ is equal to $\dbinom n k$, since we are free to choose the $k$ steps right in any order.

We can also count the number of $n$-step paths from the point $\left({k, n - k}\right)$ to $\left({n, n}\right)$.

These paths will be composed of $n-k$ steps to the right and $k$ steps up.

Therefore the number of these paths is equal to $\dbinom n {n - k} = \dbinom n k$.

Thus the total number of paths from $\left({0, 0}\right)$ to $\left({n, n}\right)$ that pass through $\left({k, n - k}\right)$ is equal to the product of:
 * the number of possible paths from $\left({0, 0}\right)$ to $\left({k, n - k}\right)$, which equals $\dbinom n k$

and:
 * the number of possible paths from $\left({k, n - k}\right)$ to $\left({n, n}\right)$, which equals $\dbinom n k$.

So the total number of paths through $\left({k, n - k}\right)$ is equal to $\dbinom n k^2$.

Summing over all possible values of $k \in 0, \ldots, n$ gives the total number of paths.

Thus we get:
 * $\displaystyle \sum_{k \mathop = 0}^n \binom n k^2 = \binom {2 n} n$