On a proof of a binomial coefficient identity In his Integers, polynomials, and rings: a course in algebra, Ronald Irving presents the following "alternating sum" identity (Theorem 8.2 (3), p. 118):
$$
{n \choose 0} - {n \choose 1} + {n \choose 2} - \cdots + (-1)^{n-1} {n \choose n - 1} + (-1)^n {n \choose n} = 0,
$$
and, immediately after (in Exercise 8.5, (3)), he asks the reader to prove it.
I have no problem coming up with proofs for this expression.  What stumps me is Irving's hint to prove it, namely to use the fundamental recurrence
$$
{n \choose r} = {n - 1 \choose r - 1} + {n - 1 \choose r},
$$
for all integers $0 \leq r \leq n$, where $n$ is a positive integer.
Now, this fundamental recurrence is the subject of Theorem 8.1, which appears not long before (p. 117) the alternating sum identity.  IOW, the book has not given much else beyond the fundamental recurrence for the proof.
Therefore, I'm looking for a proof of the alternating sum identity that is based only on the fundamental recurrence.
NB: As I said before, I know several other proofs of the alternating sum identity.  Therefore, my question is not asking for a proof of it, but rather it asks how to use the fundamental recurrence as the basis for such a proof.
 A: $$
\begin{eqnarray}
\sum_{k=0}^n (-1)^k \binom{n}{k} & = & \binom{n}{0} + (-1)^n\binom{n}{n} + \sum_{k=1}^{n-1} (-1)^k \left(\binom{n-1}{k-1} + \binom{n-1}{k} \right) \\
& = & 1+(-1)^n + \sum_{k=1}^{n-1} (-1)^k \binom{n-1}{k} - \sum_{k=0}^{n-2} (-1)^k \binom{n-1}{k} \\
\end{eqnarray}
$$
The last two sums cancel except for the end terms, leaving: 
$$\begin{eqnarray}
& = & 1+(-1)^n + (-1)^{n-1}\binom{n-1}{n-1} - \binom{n-1}{0} \\
& = & 0
\end{eqnarray}
$$
A: It is proof 1 of this link, where the "fundamental recurrence" is called "Pascal's Rule": https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_Binomial_Coefficients_for_Given_n
Quoting their proof (without the details):
$$\displaystyle \sum_{i \mathop = 0}^n \left({-1}\right)^i \binom n i= \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \left({-1}\right)^i \binom n i + \left({-1}\right)^n \binom n n=\displaystyle \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \left({-1}\right)^i \left({\binom {n - 1} {i - 1} + \binom {n - 1} i}\right) + \left({-1}\right)^n \binom n n=\displaystyle \binom n 0 - \binom {n - 1} 0 + \left({-1}\right)^{n - 1} \binom {n - 1} {n - 1} + \left({-1}\right)^n \binom n n=0$$ as a telescoping series.
