Let me work out Lucia's comment so that this question can be closed. On the one hand, by the beta integral we have $$ \int_0^1 (1-x)^m x^{n-1}\,dx = \frac{m!(n-1)!}{(m+n)!}=\frac{1}{n\binom{m+n}{n}}.$$ The same integral equals, by the binomial theorem, $$ \int_0^1 (1-x)^m x^{n-1}\,dx = \int_0^1 \sum_{k=0}^m \binom{m}{k}(-1)^kx^{n+k-1}\,dx= \sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{n+k}.$$ Hence $$ \sum_{k=0}^m\binom{m}{k}\frac{(-1)^k}{n+k}=\frac{1}{n\binom{m+n}{n}},$$ and the result follows.

This is an elaboration of Lucia's comment. On the one hand, by the beta integral we have $$ \int_0^1 (1-x)^m x^{n-1}\,dx = \frac{m!(n-1)!}{(m+n)!}=\frac{1}{n\binom{m+n}{n}}.$$ The same integral equals, by the binomial theorem, $$ \int_0^1 (1-x)^m x^{n-1}\,dx = \int_0^1 \sum_{k=0}^m \binom{m}{k}(-1)^kx^{n+k-1}\,dx= \sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{n+k}.$$ Hence $$ \sum_{k=0}^m\binom{m}{k}\frac{(-1)^k}{n+k}=\frac{1}{n\binom{m+n}{n}},$$ and the result follows.