Julia Set of polynomials If $f$ is a polynomial and $z\in\mathbb{C}$, show that either $f^n(z)\rightarrow\infty$ or $\{f^n(z) : n\geq 1\}$ is a bounded set.
Here, $f^2(z)=f(f(z))$ and $f^n(z)=f(f^{n-1}(z))$ for $n\geq 2$

I had a proof structured as followed:
1) Suppose $\{f^n(z) : n\geq 1\}$ is unbounded. Then there exists a subsequence $(n_k)$ such that $f^{n_k}(z)\rightarrow\infty$.
2) We are done if we can show that $|f^{k}(z)|$ is monotone everntually.
But the trouble is 2) is really tedious to verify. I am just wondering whether there is a more pretty way to do this.
 A: I will not treat the case where $f$ is of degree at most 1.
So assume that $f$ is a polynomial of degree at least 2.
Then $|f(z)|$ grows faster than $|z|$ as $|z|\to +\infty$. This means that there is some $M>0$ such that for all $z\in\mathbb{C}$ with $|z|>M$ we have
$$|f(z)|\geq |z|.$$
(Try to make this precise if you have doubts.)
We can conclude that as soon as the sequence gets further than $M$ from the origin, it will only get even further from the origin. This proves the result.
Remark.
We have indeed shown that if $\{f^n(z)\}$ is unbounded, then $|f^k(z)|$ is monotone eventually. Note that you could strengthen the argument above to show that if $\{f^n(z)\}$ is unbounded, then $|f^k(z)|$ will eventually grow at least exponentially (for example, take $M$ big enough to make sure that $|f(z)|\geq 2|z|$).
A: Here's an alternative proof.  I'm certain that it's not the proof expected in an elementary class, as the tools it uses are a bit heavier.  It has the advantage, though, of introducing the idea of conjugation and how it can be used to treat the point at $\infty$ - an important step towards extending polynomial dynamics to the dynamics of rational functions on the Riemann sphere.
The idea is to replace the polynomial $f(z)$ with the rational function $F(z)=1/f(1/z)$ and study the orbit of $F$.  It turns out that $F$ has a super-attractive fixed point at zero and the statement about orbits of $f$ being attracted to $\infty$ can be recast in terms of orbits of $F$ being attracted to zero.
More precisely, let $f$ be a complex polynomial of degree at least two and define 
$$F(z) = \left\{\begin{array}{ll}
  1/f(1/z) & z \neq 0 \\
  0 & z=0.
\end{array}\right.
$$
Essentially, we've simply conjugated by the reciprocal function and filled the resulting removable discontinuity at the origin.  $F$ is, in fact, differentiable at the origin with $F'(0)=0$, since
$$F'(0)=\lim_{z\rightarrow 0} \frac{F(z)-F(0)}{z-0} = \lim_{z\rightarrow 0} \frac{1}{zf(1/z)}=0.$$
That last equality is true because $f$ has degree at least two and, thus, $f(1/z)\rightarrow \infty$ faster than $z\rightarrow 0$.
Next, note that 
$$F(F(z)) = 1/f (1/(1/f(1/z))) = 1/f(f(1/z))$$
and, more generally,
$$F^n(z)=1/f^n(1/z).$$
Substituting $z$ for $1/z$, we get
$$F^n(1/z)=1/f^n(z).$$
As a result, an orbit of $f$ generates an orbit of $F$ by taking the reciprocals of the terms in the orbit and vice-versa.  Thus, an orbit in the neighborhood of $\infty$ for $f$ can be treated as an orbit in the neighborhood of $0$ for $F$.
We can now establish the result using the fact that zero is an attractive fixed point for $F$, for then there is an $r>0$ such that $|z|<r$ implies that $z$ converges to zero under iteration of $F$.  Thus, if $|z|>1/r$, then $z$ converges to $\infty$ under iteration of $f$.
Note that the number $1/r$ becomes the crucial bound in the original question.
