Prove that $\oint_{|z|=r} {dz \over P(z)} = 0$

Using the ML inequality:

$$\left|\oint_{|z|=R}\frac{dz}{p(z)}\right|\le2\pi R\cdot\max_{|z|=R}\frac1{|p(z)|}\le2\pi R\frac1{R^n}\xrightarrow[R\to\infty]{}0$$

since $\;n\ge 2\; $ .

Why? Because of the maximum modulus principle:

$$p(z)=\sum_{k=0}^na_kz^k=z^n\sum_{k=0}^na_kz^{k-n}\stackrel{\forall\,|z|=R}\implies\left|p(z)\right|\ge|z|^n\left(\left|a_n\right|-\left|\frac{a_{n-1}}z\right|-\ldots-\left|\frac{a_0}{z^n}\right|\right)\ge |a_n|R^n$$

the last equality being true for $\;R\;$ big enough since the expression within the parentheses tends to $\;|a_n|\;$ .