A common problem in Combinatorial Analysis Please help me prove the fact below:
$$
\sum_{n=1}^N \frac{N!}{n!(N-n)!}\frac{(-1)^{n-1}n}{n+x} = \frac{N!}{\prod_{n=1}^N (n+x)}.
$$
I think this problem is common, but it is really hard for me to solve.

Alternatively to Jack's answer, multiply both sides by $\dfrac{\prod_{n=1}^N (x+n)}{N!}$ and you get an equality of polynomials of degree at most $N$, and show that the equality is true for $x=0,-1,-2,\dots,-N$. 
The cases $x=-1,\dots,-N$, the left hand side is zero for all but one term. which you can check easily.
The case $x=0$ reduces to the binomial theorem for $(1+(-1))^N$, specifically:
$$1-\left(1+(-1)\right)^N = 1$$
A possible inductive proof would write the left hand side as $f_N(x)$ and then you'd get that you want to show that:
$$f_N(x) = \frac{N}{N+x} f_{N-1}(x)$$
