Solution:
We claim that the sum of the squares of the numbers in row $n$ is
$$\left(\begin{array}{l} n \\ 0 \end{array}\right)^2+\left(\begin{array}{c} n \\ 1 \end{array}\right)^2+\cdots+\left(\begin{array}{l} n \\ n \end{array}\right)^2=\left(\begin{array}{c} 2 n \\ n \end{array}\right)$$
the middle number in row $2 n$ of Pascal's triangle. To prove this, we can think of the right side as the number of ways to form a committee of $n$ people chosen from some $2 n$ people. For the left side, suppose the $2 n$ people are $n$ men and $n$ women. If there are $k$ women on the committee, then there must be $n-k$ men, so there are $\left(\begin{array}{l}n \\ k\end{array}\right)$ ways to choose the women and $\left(\begin{array}{c}n \\ n-k\end{array}\right)$ ways to choose the men, giving $\left(\begin{array}{c}n \\ k\end{array}\right)\left(\begin{array}{c}n \\ n-k\end{array}\right)=\left(\begin{array}{l}n \\ k\end{array}\right)^2$ possible committees with $k$ women. Adding over the possible values of $k$ from 0 to $n$ gives the sum on the left. Since each side counts the same thing in a different way, the two sides must be equal.