Let $n$ be a positive integer, Prove that $\sum_{k=1}^n\frac{ (-1)^{k-1}}{k}{n \choose k} = H_n$

Question

Let $n$ be a positive integer, Prove that $\sum_{k=1}^n \frac{(-1)^{(k-1)}}{ k} {n \choose k} = H_n = 1+\frac{1}{2}+\ldots+\frac{1}{n}.$

This problem was solved as an example in Titu Andreescu's Combinatorics book. However I do not understand the last part of the solution. Here is the solution given by him

I was able to understand the solution presented above, however. What I do not understand in this

For those who might want to ask what theorem 3.2(i) states, here it is

Please someone should help me demystify this solution, Thanks in advance.

I'm not too good with Mathjax so here is the screenshot of the original problem.

Answer 1

He shows that $$ \sum_{k=0}^{m+1}\left(-1\right)^{k}\binom{m+1}{k}=0 $$ He uses the following property $(i)$ from the theorem $3.2$