Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 2

Proof
We have by definition of vacuous summation that:
 * $\ds \forall n \in \Z: n < 0: \sum_{i \mathop = 0}^n \binom n 1 = 0$

Then from Zero Choose Zero:
 * $\ds \sum_{i \mathop = 0}^0 \binom 0 0 = 1$

Let $n > 0$.

From the Binomial Theorem, we have that:


 * $\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i} y^i$

Putting $x = 1, y = -1$, we get: