First, it can be shown that $$\frac{1}{k}\binom{n}{k} = \frac{1}{k}\left(\binom{n-1}{k} +\binom{n-1}{k-1}\right)=\frac{1}{k}\binom{n-1}{k} +\frac{1}{n}\binom{n}{k}$$

Let $f(n)$ be the left side of the to-be-proved identity, we are going to show the following recursion holds $$f(n)=f(n-1)+\frac{1}{n}$$

This recursion holds because $$\begin{align*} f(n) =\ &\binom{n}{1}- \frac{1}{2}\binom{n}{2}+\cdots + (-1)^{n+1}\frac{1}{n}\binom{n}{n}\\ \\=\ &\left(\binom{n-1}{1}+\frac{1}{2}\binom{n}{1}\right) -\left(\frac{1}{2}\binom{n-1}{2}+\frac{1}{3}\binom{n}{2}\right)+\cdots  +(-1)^{n+1}\left(\frac{1}{n}\right)\\ \\=\ &\left(\binom{n-1}{1}-\frac{1}{2}\binom{n-1}{2}+\cdots+(-1)^n\frac{1}{n-1}\binom{n-1}{n-1}\right)\\ \ &+\frac{1}{n}\left(\binom{n}{1}-\binom{n}{2}+\cdots+(-1)^{n+1}\binom{n}{n}\right)\\ \\=\ &f(n-1) + \frac{1}{n} \end{align*}$$

Then, it follows $$\begin{align*} f(n)=\ &f(n-1)+\frac{1}{n}\\ \\=\ &f(n-2)+\frac{1}{n-1}+\frac{1}{n}\\=\ &\cdots\\=\ &f(1)+\frac{1}{2}+\cdots + \frac{1}{n-1}+\frac{1}{n}\\ \\=\ &1+\frac{1}{2}+\cdots + \frac{1}{n-1}+ \frac{1}{n} \end{align*}$$