Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 3

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 $n > 0$.

The assertion can be expressed:
 * $\ds \sum_{i \mathop \le n} \paren {-1}^i \binom n i = 0$ for all $n > 0$

as $\dbinom n i = 0$ when $i < 0$ by definition of binomial coefficient.

From Alternating Sum and Difference of r Choose k up to n we have:
 * $\ds \sum_{i \mathop \le n} \paren {-1}^i \binom r i = \paren {-1}^n \binom {r - 1} n$

Putting $r = n$ we have:
 * $\ds \sum_{i \mathop \le n} \paren {-1}^i \binom n i = \paren {-1}^n \binom {n - 1} n$

As $n - 1 < n$ it follows from the definition of binomial coefficient that:
 * $\dbinom {n - 1} n = 0$