Sum of Squares of Binomial Coefficients/Algebraic Proof

Proof
Consider the Binomial Theorem:

Let $m$ be the coefficient of $x^n$ in the expansion of $\paren {1 + x}^{2 n} = \paren {1 + x}^n \times \paren {1 + x}^n$.

For all $b \in \set {0, 1, n}$, by matching up the coefficient of $x^k$ with that of $x^{n - k}$ in $(1)$, we have:
 * $m = \ds \sum_{k \mathop = 0}^n \dbinom n k^2$

But from the Binomial Theorem, this given by:
 * $m = \dbinom {2 n} n$

Hence the result.