Binomial coefficient identity It seems to be nontrivial (to me) to show that the following identity holds:
$$ \binom {m+n}{n} \sum_{k=0}^m \binom {m}{k} \frac {n(-1)^k}{n+k} = 1. $$
This quantity is related to the volume of the certain polytope.

This is a specialization at $x = n$ of the rational function identity
$$
\frac{1}{x(x+1)\cdots(x+m)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{x+k}, 
$$
and this identity could be proved either by partial fractions with unknown coefficients and limits to find the coefficients (see my comment on Todd Trimble's answer) or by induction on $m$. For the inductive step, divide both sides of the above identity by $x+m+1$:
$$
\frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(x+k)(x+m+1)}.
$$
Split up the right side by partial fractions to make it 
$$
\frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{m+1-k}\left(\frac{1}{x+k} - \frac{1}{x+m+1}\right).
$$
Break this into a sum of two terms:
$$
\frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{1}{m+1-k}\frac{(-1)^k}{x+k} - \frac{1}{m!}\left(\sum_{k=0}^m\binom{m}{k}\frac{(-1)^k}{m+1-k}\right)\frac{1}{x+m+1}.
$$
Since $\binom{m}{k}/(m+1-k) = \binom{m+1}{k}/(m+1)$, the difference is
$$
\frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\left(\sum_{k=0}^m\binom{m+1}{k}(-1)^k\right)\frac{1}{x+m+1}.
$$
The sum inside parentheses is $(1-1)^{m+1} - (-1)^{m+1} = (-1)^m$, so finally we have
$$
\frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\frac{(-1)^m}{x+m+1}.
$$
Since $-(-1)^m = (-1)^{m+1}$, we're done.
