If you do possess Concrete Mathematics, check out 5.1 Basic Identities and 5.2 Basic Practice. The list can be found at P174 Table 174 The top ten binomial coefficient identities.

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 then calculate 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 · $$ · \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\left(\sum_{k=0}^m\binom{m+1}{k}(-1)^k\right)\frac{1}{x+m+1}. · $$ · In the second term, the sum inside parentheses is $(1-1)^{m+1} - (-1)^{m+1} = (-1)^m$, so finally we have · $$ · \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\frac{(-1)^m}{x+m+1}. · $$$-(-1)^m = (-1)^{m+1}$, we're done.