Proving $n - \frac{_{2}^{n}\textrm{C}}{2} + \frac{_{3}^{n}\textrm{C}}{3} - ...= 1 + \frac{1}{2} +...+ \frac{1}{n}$ Prove that $n - \frac{_{2}^{n}\textrm{C}}{2} + \frac{_{3}^{n}\textrm{C}}{3} - ... (-1)^{n+1}\frac{_{n}^{n}\textrm{C}}{n} = 1 + \frac{1}{2} + \frac{1}{3} +...+ \frac{1}{n}$
I am not able to prove this. Please help!
 A: 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.
A: @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.
