Proving $n - \frac{_{2}^{n}\textrm{C}}{2} + \frac{_{3}^{n}\textrm{C}}{3} - ...= 1 + \frac{1}{2} +...+ \frac{1}{n}$

Hint. One may observe that $$ \frac1k=\int_0^1t^{k-1}dt,\quad k\geq1, $$ giving $$ \sum_{k=1}^n\frac1k=\int_0^1\sum_{k=1}^nt^{k-1}dt=\int_0^1\frac{1-t^{n}}{1-t}dt=\int_0^1\frac{1-(1-u)^{n}}udu $$ then use the binomial theorem in the latter integrand to conclude.

@Olivier Oloa's proof is so brilliant! Even though, we can still prove this by induction on $n$. For $n=1$, then the result is clear. Now, suppose that the result holds for some $n\in\mathbb{N}$, then for $n+1$, we have \begin{align} \sum_{k=1}^{n+1}\frac{(-1)^{k+1}}{k}{n+1\choose k} &=\sum_{k=1}^n\frac{(-1)^{k+1}}{k}\cdot\frac{n+1}{n-k+1}{n\choose k}+\frac{(-1)^{n+2}}{n+1}\\ &=\sum_{k=1}^n\frac{(-1)^{k+1}}{k}{n\choose k} +\sum_{k=1}^n\frac{(-1)^{k+1}}{n-k+1}{n\choose k} -\frac{(-1)^{n+1}}{n+1}\\ &=\sum_{k=1}^n\frac{1}{k} -\frac{1}{n+1}\sum_{k=1}^n\frac{(-1)^{k}(n+1)}{n-k+1}{n\choose k} -\frac{(-1)^{n+1}}{n+1}\\ &=\sum_{k=1}^n\frac{1}{k} -\frac{1}{n+1}\sum_{k=0}^{n+1}(-1)^{k}{n+1\choose k} +\frac{1}{n+1}\\ &=\sum_{k=1}^n\frac{1}{k}-\frac{(1-1)^{n+1}}{n+1}+\frac{1}{n+1}\\ &=\sum_{k=1}^{n+1}\frac{1}{k}. \end{align} This completes the proof.