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!

Another solution can be based on the change of variable $z = \frac{R^{2}}{w}$. 
If
$$
p(z) =cz^{m}\prod_{k=1}^{n-m}(z-a_{k})
$$
where $n \ge 2$, $0 \le m \le n$ and $c \neq0$ is a constant, then
$$
p\left(\dfrac{R^{2}}{w}\right) = cR^{2m}w^{-n}\underbrace{\prod_{k=1}^{n-m}(R^{2}-a_{k}w)}_{= q(w)} = cR^{2m}\dfrac{q(w)}{w^{n}}.
$$
Futhermore, if $0 < |a_{k}|<R$ then $\left|\frac{R^{2}}{a_{k}}\right| > R.$
We are now prepared to change the variable $z$ to $w = \frac{R^{2}}{z}$. The circle $|z| = R$ will be transformed to the circle $|w| = R$ but with a reversed orientation. Thus
$$
\oint_{|z|=R}\dfrac{1}{p(z)}\, dz = -\oint_{|w|=R}\dfrac{w^{n}}{cR^{2m}q(w)}\cdot\dfrac{-R^{2}}{w^{2}}\, dw = \dfrac{R^{2-2m}}{c}\oint_{|w|=R}\dfrac{w^{n-2}}{q(w)} \, dw.
$$
Since $n \ge 2$ and $q(w)$ has no zeros inside the circle $|w| = R$
 the last integral will be $0$.
