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.
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 $|f(z)|\geq 2|z|$).