If $f$ has an essential singularity at $0$, there is a sequence $z_n \to 0$ such that $z_n^n f(z_n) \to \infty$ Here's a problem I was just working on:

Let $f$ have an essential singularity at $0$.  Show that there is a sequence of points $z_n \to 0$ such that $z_n^n f(z_n)$ tends to infinity.

I know already that there exists a sequence $z_n \to 0$ such that $f(z_n)$ tends to any complex number I want, hence I can get a sequence that tends to infinity.  The problem is that I need this sequence to tend to infinity really fast.  
What I did so far was look at the function $g_n(z) := z^n f(z)$, $n \geq 1$.  This obviously also has an essential singularity at $0$, so I can find a sequence $z_{n1}, z_{n2}, ...$ such that $\lim\limits_k g_n(z_{nk}) = \infty$.  Do you think it's possible to extract from the array $z_{nk}, (n,k) \in \mathbb{N}^2$ a subsequence $z_l$ such that $g_l(w_l) \to \infty$ as $l \to \infty$?  I tried for awhile but I'm just not very good at these kinds of arguments.
 A: Yes. It is possible, but I think it will be easier to rebuild it from scratch than to extract such a sequence.
You can define $z_n$ in the following way:
For each $n$, there exists $z_n \in B(0,1/n)$ such that $|g_n(z_n)| > n$ (since, as you said yourself, there is a sequence that tends to zero and takes $g_n$ to $\infty$). This is your sequence. It should be easy to finish the proof from here.
A: You should be able to extract such a sequence from $g_n(z_{n,k})$. To put this sequence property slightly differently, you know that for any $n \in \mathbb{N}$, any complex number $w$, and any $\varepsilon > 0$, you can find a complex number $z$ such that $|g_n(z) - w| < \varepsilon$. So, fix $n \in \mathbb{N}$, choose $\varepsilon = 1$, and $w = n$, and you get a sequence of corresponding $z_n$'s satisfying,
$$|g_n(z_n) - n| < 1.$$
Basically, $g_n(z_n)$ is a sequence that's always within distance $1$ of $n$, so it shouldn't tend to anything other than $\infty$. To prove this, use reverse triangle inequality. We have,
$$||g_n(z_n)| - n| \le |g_n(z_n) - n| < 1,$$
so
$$-1 + n < |g_n(z_n)| < 1 + n.$$
By the squeeze theorem, $|g_n(z_n)| \rightarrow \infty$.
