$p$ polynomial, $p^{[n]}=n$th iterate of $p$. Set $K=\{z:p^{[n]}(z)\not\to\infty\}$. Then $J=\{z:\{p^{[n]}\}$ not normal near $z\}=\partial K$. I'm working on another exercise from a qualifying exam at my university, stated as follows:

Let $f$ be a polynomial of degree greater than 1. Let $K(f)=\{z\in\mathbb{C}\,:\,f^{[n]}(z)\not\to\infty\}$, where $f^{[n]}$ is the $n$th iterate of $f$. Then $K(f)$ is a compact subset of $\mathbb{C}$. Let $J(f)$ be the boundary $K(f)$. Show that:
$$
J(f)=\left\{z\in\mathbb{C}\,:\,\text{the family }\left\{f^{[n]}\right\}\text{ is not normal on a neighborhood of }z\right\}.
$$
Note: The constant function $\infty$ is an allowed locally uniform limit of a normal family.

I almost have a full solution but I'm stuck on the very last step.
What I have so far
Let $z\in J(f)$ and suppose that the family $\left\{f^{[n]}\right\}$ is normal on every neighborhood of $z$. Montel's theorem asserts that either $\{f^{[n]}\}$ is locally uniformly bounded on $\mathbb{C}$, or that there exists a subsequence $\{f^{[n_k]}\}$ such that $\lim_{k\to\infty}f^{[n_k]}=\infty$ uniformly on compact sets of $\mathbb{C}$. By choosing $z'\not\in K(f)$ we see that $f^{[n]}(z')\to\infty$ and hence it must be the case that such a subsequence exists. However, the (nontrivial) polynomial $f(z)-z$ has a zero in $\mathbb{C}$, and so $f$ has a fixed point $a$, whence $\lim_{n\to\infty}f^{[n]}(a)=a$. Therefore there must exist a neighborhood of $z$ on which the family $\{f^{[n]}\}$ is not normal, i.e.
$$
J(f)\subseteq\left\{z\in\mathbb{C}\,:\,\text{the family }\left\{f^{[n]}\right\}\text{ is not normal on a neighborhood of }z\right\}.
$$
Now suppose that $z$ is an element of the latter set above and let $U$ be a neighborhood of $z$. By shrinking $U$ if necessary, we may assume that $\{f^{[n]}\}$ is \textit{not} normal on $U$, so we know two things: (1) that $f^{[n]}$ is not locally uniformly bounded on $U$ (Montel's theorem), and (2) there does not exist a subsequence of $\{f^{[n]}\}$ converging locally uniformly to $\infty$ on $U$. (2) shows that $U\cap K(f)\neq\varnothing$, otherwise $f^{[n]}(w)\to\infty$ for all $w\in U$, i.e. a clear contradiction of (2). From (1), we know that there exists a $z''\in V$ such that $\lim_{k\to\infty}f^{[n_k]}(z'')\to\infty$ for some subsequence of $\{f^{[n]}\}$. If $f^{[n]}(z'')\not\to\infty$, then there exists another subsequence such that $\lim_{\ell\to\infty}f^{[n_\ell]}(z'')=b<\infty$. I do not know where to go from here.
$$
$$
I've also tried writing $f$ out as $f(z)=c_1(z-a)+c_2(z-a)^2+\ldots+c_n(z-a)^n$, where $a$ is a root of $f$. This has given me a bit of insight but nothing incredibly useful.
Thoughts
I do not feel like I've adequately used the fact that $f$ is a polynomial of degree greater than 2 here, so I would guess I'm overlooking some fact about polynomials. I also feel like my approach is a bit clumsy, and there's probably a better way of going about this problem.
$$
$$
I'd also like to quickly apologize for posting so many questions lately. I'm hosting a prep course for the qualifying exam in complex analysis, and I'd just like to be able to show complete solutions for the students preparing. Thanks a bunch.
 A: I'll answer my own question after thinking about it for a while.
Solution
Let $z\in J(f)$. Then in any neighborhood $U$ of $z$ we may find points $z'\in U$ such that $f^{[n]}(z')\to\infty$, however $z$ itself has $f^{[n]}(z)\not\to\infty$. Therefore $\{f^{[n]}\}$ cannot be normal in any neighborhood of $z$, and hence
$$
J(f)\subseteq\left\{z\in\mathbb{C}\,:\,\text{the family }\left\{f^{[n]}\right\}\text{ is not normal on a neighborhood of }z\right\}.
$$
Conversely, let $z$ be a member of the latter set above. If $z\not\in K(f)$, then there exists an open neighborhood $V$ about $z$ such that $V\subseteq\mathbb{C}\setminus K(f)$, as $K(f)$ is compact. In other words, $f^{[n]}(w)\to\infty$ for all $w\in V$.
We claim that $f^{[n]}\to\infty$ uniformly on compact subsets of $V$ and hence is normal. Accordingly, let $L$ be a compact subset of $V$. Because $f$ is at least quadratic, there exists a $R$ such that $|f(z)|>|z|$ for $|z|>R$. Fix $M>R$ and set $E_n=\{z\in L\,:\, |f^{[n]}(z)|>M\}$. Each $f^{[n]}$ is continuous, and so each $E_n$ is open. Furthermore, if $w\in E_n$, then
$$
|f^{[n+1]}(z)|\geq |f^{[n]}(z)|> M>R
$$
and $w\in E_{n+1}$. Thus $\{E_n\}$ is an ascending sequence; that is, $E_1\subseteq E_{2}\subseteq E_{3}\subseteq\ldots$. Since $f^{[n]}\to\infty$ pointwise on $L$, $\{E_n\}$ is an open cover of $L$. By compactness there is a finite subcover of $\{E_n\}$, and hence there is a positive integer $N$ such that $E_N=L$; in other words $|f^{[n]}(z)|>M$ for all $n\geq N$ and $z\in L$. We conclude that $f^{n}\to\infty$ uniformly on $L$ and thus $z\in K(f)$ by contradiction.
Now suppose that there exists a  neighborhood $W$ of $z$ that is contained in $K(f)$. Because $f$ is at least quadratic, we know that there exists a $R$ such that $|f(z)|>|z|$ for $|z|> R$. For each $w\in W$, $\limsup_{n}f^{[n]}(w)\leq R$, otherwise $f^{[n_0]}(w)>R$ for some $n_0$ and hence $f^{[n]}(w)\to\infty$. Therefore $\{f^{[n]}\}$ is bounded on $W$, and hence normal by Montel's theorem; this is a contradiction. We conclude that every neighborhood of $z\in K(f)$ contains a member of $\mathbb{C}\setminus K(f)$, and hence
$$
J(f)\supseteq\left\{z\in\mathbb{C}\,:\,\text{the family }\left\{f^{[n]}\right\}\text{ is not normal on a neighborhood of }z\right\}.
$$
Therefore the two sets are equal.
