Essential singularities of holomorphic functions and a certain limit Let $f$ be analytic in the punctured unit disc in $\Bbb C$, and suppose $0$ is an essential singularity for $f$. Let $M_{f}(r) = \max_{|z|=r}|f(z)|$. Prove that
$$\lim_{r \to 0^{+}}r^{n}M_{f}(r)=\infty$$. 
for all $n \in \Bbb N$.
So I think the idea is to analyze the Laurent series expansion for $f$, and to use the fact that infinitely many negative powers exist in the expansion. Where do I go from here?
EDIT: Specifically, I am trying to use Cassorati-Weierstrass to find a sequence convergence to 0 in the punctured disc whose image grows as a sequence that blows up faster than r^n shrinks (say, n fixed). I am having trouble formulating this notion.
 A: We'll show that if $r^n M_f(r) \not \to \infty$ then the function $z^n f(z)$ can be holomorphically extended at $0$. 
Assume that 
$$r^{n}M_{f}(r)\not \to \infty$$
Thus, there exists a constant $C>0$ and a sequence of positive numbers $r_k \to 0$ so that $$r_k^n M_{f}(r_k) \le C$$
Then for the function $g(z) = z^{n} f(z)$  we have 
$$M_g(r_k) \le C $$
Let's note now that if for a holomorphic $g$ on the pointed disk we have 
$M_g(r_k) \le C$ for a sequence $r_k \to 0$ then the singularity $0$ is removable, that is $g$ can be extended holomorphically the the disk. Indeed, consider $z \in D$, $z\ne 0$. Consider $r_k < |z|$. By the Cauchy formula we have 
$$ g(z) = \frac{1}{2 \pi i}\left( \int_{C_1} \frac{g(\zeta)}{ \zeta - z}-\int_{C_{r_k}} \frac{g(\zeta)}{ \zeta - z} \right)$$
Note that for $k\to \infty$ the term $\displaystyle\int_{C_{r_k}} \frac{g(\zeta)}{ \zeta - z}\to 0$ . Indeed, 
$$ \mid \int_{C_{r_k}} \frac{g(\zeta)}{ \zeta - z}\mid \,\le \, 2 \pi r_k\cdot  M_g(r_k) \cdot \frac{1}{|z| - r_k} $$
We conclude that 
$$ g(z) = \frac{1}{2 \pi i}\cdot  \int_{C_1} \frac{g(\zeta)}{ \zeta - z} $$
for all $z$ in the pointed disk and therefore $g$ can be extended to the whole disk.
