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

Proof
The assertion can be expressed:
 * $\displaystyle \sum_{i \mathop \le n} \left({-1}\right)^i \binom n i = 0$ 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:
 * $\displaystyle \sum_{i \mathop \le n} \left({-1}\right)^i \binom r i = \left({-1}\right)^n \binom {r - 1} n$

Putting $r = n$ we have:
 * $\displaystyle \sum_{i \mathop \le n} \left({-1}\right)^i \binom n i = \left({-1}\right)^n \binom {n - 1} n$

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