Binomial theorem relating proof There is this identity 
$$1 -\frac{1}{2}\binom{n}{1}+\frac{1}{3} \binom{n}{2}- \frac{1}{4}\binom{n}{3}+....+(-1)^n \frac{1}{n+1}\binom{n}{n}$$
And we are supposed to prove it using these two identities 
$$k\binom{n}{k} = n\binom{n-1}{k-1}$$
and 
$$\binom{n}{0} + \binom{n}{1} + \binom{n}{2} +....+ \binom{n}{n} = 2^n$$
I have been working on this problem for a long time.  Can you guys help me?

\begin{eqnarray*}
\frac{1}{i+1}= \int_0^1 x^i dx
\end{eqnarray*}
Sub this into the sum & interchange the order of the sum & the integral
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^i\binom{n}{i} \frac{1}{i+1}&=& \int_0^1  \sum_{i=0}^{n} (-1)^i\binom{n}{i}x^i dx \\
&=& \int_0^1  (1-x)^n dx \\
&=&  \left[  \frac{-(1-x)^n}{n+1} \right]^1_0 \\
&=&   \color{blue}{ \frac{1}{n+1}}  \\
\end{eqnarray*}
