Prove that the identity is true for all natural numbers for the identity:
$$\frac{n!}{x(x+1)(x+2)...(x+n)} = \frac{A_0}{x+0} + \frac{A_1}{x+1}+...+ \frac{A_n}{x+n}$$
prove
$$A_k= (-1)^kC(n,k)$$
I think this might work by induction, but i am not able to arrive at a final answer. please help!!!

As what you tagged, we prove it by induction. For $n=0$, then the result is 
trivial. Suppose that the result holds for some $n\ge 0$, then for $n+1$,
\begin{align}
&\frac{(n+1)!}{x(x+1)(x+2)\cdots[x+(n+1)]}\\
&\quad\quad=\frac{n!}{x(x+1)(x+2)\cdots(x+n)}\cdot\frac{n+1}{x+n+1}\\
&\quad\quad=\left(\frac{A_0}{x+0}+\frac{A_1}{x+1}+\cdots+\frac{A_n}{x+n}\right)\cdot\frac{n+1}{x+n+1}\\
&\quad\quad=\frac{(n+1)A_0}{(x+0)(x+n+1)}+\frac{(n+1)A_1}{(x+1)(x+n+1)}+\cdots+\frac{(n+1)A_n}{(x+n)(x+n+1)}\\
&\quad\quad=\left[\frac{(n+1)A_0}{x+0}-\frac{(n+1)A_0}{x+n+1}\right]\cdot\frac{1}{n+1}
+\left[\frac{(n+1)A_1}{x+1}-\frac{(n+1)A_1}{x+n+1}\right]\cdot\frac{1}{n}
\\&\quad\quad\,\,\,\,\,\,\,+\cdots+\left[\frac{(n+1)A_n}{x+n}-\frac{(n+1)A_n}{x+n+1}\right]\\
&\quad\quad=\frac{A_0}{x+0}+\frac{\frac{n+1}{n}A_1}{x+1}+\cdots+\frac{(n+1)A_n}{x+n}
-\frac{A_0+\frac{n+1}{n}A_1+\cdots+(n+1)A_n}{x+n+1}\\
&\quad\quad=\frac{A_0'}{x+0}+\frac{A_1'}{x+1}+\cdots+\frac{A_n'}{x+n}
+\frac{A_{n+1}'}{x+n+1},
\end{align}
where $A_0',A_1',\ldots,A_{n+1}'$ denotes the new coefficient in terms of the case $n+1$. Now, by the assumption, we check that
\begin{align*}
A_k'=\frac{n+1}{n-k+1}A_k=\frac{n+1}{n-k+1}(-1)^k{n\choose k}=(-1)^k{n+1\choose k}\quad\mbox{for }0\le k\le n,
\end{align*}
and
\begin{align*}
A_{n+1}'
&=-\left[A_0+\frac{n+1}{n}A_1+\cdots+(n+1)A_n\right]\\
&=-\sum_{k=0}^n(-1)^k\frac{n+1}{n-k+1}{n\choose k}\\
&=-\sum_{k=0}^n(-1)^{k}{n+1\choose k}\\
&=-\sum_{k=0}^{n+1}(-1)^{k}{n+1\choose k}+(-1)^{n+1}\\
&=(-1)^{n+1}.
\end{align*}
This completes the proof.
