Prove that $\oint_{|z|=r} {dz \over P(z)} = 0$ I got stuck on this problem, hope anyone can give me some hints to go on solving this:

P is a polynomial with degree greater than 1 and all the roots of $P$ in complex plane are in the disk B: $|z| = r$. Prove that: $$\oint_{|z| = r} {{dz}\over{P(z)}} = 0$$
  Here, the direction of the integral is the positive direction(actually, it can take whatever direction, because the value of the integral is 0).

What I tried so far: Applying D'Alembert-Gauss theorem, we can write $P(z) = (z-z_1)^{p_1}(z-z_2)^{p_2}...(z-z_n)^{p_n}$, here $z_i$ are complex numbers which different from each other. We can choose for each $i = 1,...,n$ a $r_i > 0$ small enough such that $B(p_i,r_i)$ are disjoint with each others and all belong to $B$. So use Cauchy theorem for compact Jordan region generated by $B$ and $B(p_i, r_i)$, it's easy to see that:
$$\oint_{|z| = r} {{dz}\over{P(z)}} = \sum_1^{n}{\oint_{|z-z_i|=r_i} {{dz}\over{P(z)}}} =  \sum_1^{n}{\oint_{|z-z_i|=r_i} {{\prod_{j \neq i}{1 \over {(z-z_j)^{p_j}}}}\over{(z-z_i)^{p_i}}}}$$. Then I tried to apply Cauchy theorem for each ${\oint_{|z-z_i|=r_i} {{\prod_{j \neq i}{1 \over {(z-z_j)^{p_j}}}}\over{(z-z_i)^{p_i}}}}$:
$$f^{(k)}(z) = {k! \over {2 \pi i}} \oint_{\partial{B}}{{f(t) \over (t-z)^{k+1}}dt}$$, here $\partial{B}$ is notion for the boundary of the disk $B$
But I got stuck when trying to calculate the $(p_i - 1)$-th derivative for ${\prod_{j \neq i}{1 \over {(z-z_j)^{p_j}}}}$. I expect that each expression should be equal to 0, but I can't prove it. Anyone has any ideas to move on? If there's any point unclear, please don't hesitate to ask me. Thanks!

Let $a_1,\ldots,a_r$ be the roots of $P(z)$, with multiplicities $m_1,\ldots,m_r$.
Consider the partial fraction decomposition of $\frac{1}{P(z)}$,
$$\frac{1}{P(z)}=\sum_{i=1}^r\sum_{j=1}^{m_i}\frac{c_{i_j}}{(z-a_i)^j}.$$
Then since $\text{deg}(P) > 1$,
$$0=\lim_{z\rightarrow\infty}\frac{z}{P(z)}=\lim_{z\rightarrow\infty}z\sum_{i=1}^r\sum_{j=1}^{m_i}\frac{c_{i_j}}{(z-a_i)^j}=\sum_{i=1}^rc_{i_1}.$$
Using this fact, and then applying Cauchy's integral formula (for derivatives) to $\int_{|z|=R}\frac{dz}{P(z)}$, we have,
$$\int_{|z|=R}\frac{dz}{P(z)}=\int_{|z|=R}\sum_{i=1}^r\sum_{j=1}^{m_i}\frac{c_{i_j}}{(z-a_i)^j}dz=\int_{|z|=R}\sum_{i=1}^r\frac{c_{i_1}}{z-a_i}dz=2\pi i\sum_{i=1}^rc_{i_1}=0.$$
QED
