If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function 
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then differentiating we get
$$\frac{f'(x)}{f(x)}=\sum_{k=0}^n \frac{A_k}{x+k}$$ 
hence
$$\frac{f'(x)}{f(x)}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$
any clue here?

Note that:
$$\frac1x\frac1{\binom{x+n}n}=\frac{n!}{x(x+1)(x+2)\dots(x+n)}$$
And thus,
$$\frac1x\left[\frac1{\binom{x+n}n}-\frac1{\binom{x+{n-1}}{n-1}}\right]=\frac{A_n}{x+n}$$
Which gives the general solution immediately.
