Here is an even simpler take on this. Let $$F(m,n)=\binom {m+n}{n} \sum_{k=0}^m \binom {m}{k} \frac {n(-1)^k}{n+k}.$$
Claim: If $m\geq 1$, then we have$$F(m,n)=F(m-1,n+1),$$ for all $n$.
proof: $$F(m,n)=\binom {m+n}{n} \sum_{k=0}^m \binom {m}{k} (-1)^k \left(1-\frac{k}{n+k}\right)=\binom {m+n}{n} \sum_{k=0}^m \binom {m}{k} \frac {k(-1)^{k+1}}{n+k}$$ where I used the fact that $\sum_{k} \binom{m}{k}(-1)^k=0$ for $m\geq 1$. We can further say $$F(m,n)=\binom{(m-1)+(n+1)}{n+1}\sum_{r=0}^{m-1}\binom{m-1}{r}\frac{(n+1)(-1)^{r}}{n+1+r},$$ where I basically just substituted $r=k-1$ and rearranged the terms. However this last sum is just $F(m-1,n+1)$.
Moreover, it is very easy to check that $F(0,m+n)=1$.