Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 1

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

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

For $n > 0$:

We note:
 * $\dbinom n 0 = \dbinom {n - 1} 0 = 1$

so:
 * $\dbinom n 0 - \dbinom {n - 1} 0 = 0$


 * $\left({-1}\right)^{n - 1} \dbinom {n - 1} {n - 1} = - \left({-1}\right)^n \dbinom n n = - \left({-1}\right)^n$

so:
 * $\left({-1}\right)^{n - 1} \dbinom {n - 1} {n - 1} + \left({-1}\right)^n \dbinom n n = 0$

Hence the result.