Square of Difference/Algebraic Proof 2

Proof
Follows directly from the Binomial Theorem:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \left({x + y}\right)^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$

putting $n = 2$ and $y = -y$.