Harmonic numbers problem

Prove that $$\sum_{i=1}^{n}\frac{(-1)^{i-1}}{i}{n\choose i}=H_n$$
 where $H_n$ is the $n$-th harmonic number.

Proof. Denotе the sum by $S$. Consider $$f(t)=\sum_{i=1}^{n}{n\choose i}t^{i-1}=\frac{(1+t)^n-1}{t}$$ Now integrate
$$\int_{0}^{x} f(t)\mathrm{d}t=\sum_{i=1}^{n}{n\choose i}\frac{x^{i}}{i}=\int_{0}^{x}\frac{(1+t)^n-1}{t}\mathrm{d}t$$ Now plug $x=-1$ and we get $$\int_{0}^{-1}f(t)\mathrm{d}t=-S$$
On the other hand $$\int_{0}^{-1}f(t)\mathrm{d}t=\int_{0}^{-1}\frac{(1+t)^n-1}{t}\mathrm{d}t=-\int_{0}^{1}\frac{s^n-1}{s-1}\mathrm{d}s=\\=-\int_{0}^{1}1+s+s^2+\cdots+s^{n-1}\mathrm{d}s=-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)=-H_n$$