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!

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.
