Prove that $ \binom nm \sum_{i=0}^{m} (-1)^i\frac{\binom mi}{n-m+1+i}=\frac{1}{1+n}$ I want to prove that
$$
\binom nm \sum_{i=0}^{m} (-1)^i\frac{\binom mi}{n-m+1+i}=\frac{1}{1+n}\\
$$
$$
\binom nm [ \frac{\binom m0}{n-m+1} +(-1)^1\frac{\binom m1}{n-m+2}+...
..+(-1)^m\frac{\binom mm}{n+1}] 
=\frac{1}{1+n}\\
$$


*

*Assume that $m=1$,then it's equal to $1/(n+1)$.
$$
\binom n1 [ \frac{1}{n} -\frac{1}{n+1}] =\frac{1}{1+n}\\
$$

*Assume that $m=2$,then it's equal to $1/(n+1)$.
$$
\binom n2 [ \frac{1}{n-1} -\frac{2}{n}+\frac{1}{n+1}] =\frac{1}{1+n}\\
$$
How to prove that for $\forall m$, this formula is equal to $1/(n+1)$？

Hint. Note that
$$x^{n-m}(1-x)^m=\sum_{i=0}^{m} (-1)^i \binom mi x^{n-m+i}$$
Now integrate with respect to $x$ over the interval $[0,1]$ and recall the definition of Beta function.
