Julia set equality proof

I'm following the book Fractal Geometry, by K. Falconer, and in Chapter 14, Theorem 14.10 he proves that $$J(f)=J_0(f)$$, for $$f$$ polynomial and $$J(f) =$$ {closure of the set of repelling periodic points of $$f$$} and $$J_0(f) =$$ {points on which the family of iterates of $$f$$ isn't normal}.
In the second part of the proof, he considers: $$K=\{w\in J_0(f);\,\exists z\neq w;\,f(z)=w\wedge f'(z)\neq0\}$$ He then tries to show $$K\subset J(f)$$ as follows: Taking $$w\in K$$, $$\exists V$$ neighbourhood of $$w$$, and $$f^{-1}:V\rightarrow \mathbb{C}$$ local inverse. He then defines: $$h_k(z) = \frac{f^k(z)-z}{f^{-1}(z)-z}$$ In $$V$$ and uses the fact that it isn't normal to conclude it must assume the value $$0$$ or $$1$$, so that there exists a periodic point $$\overline{z}\in V$$. It then affirms that implies $$w \in J(f)$$.
My question is: the last implication requires $$\overline{z}$$ to be a $$\textbf{repelling}$$ periodic point, but I can't prove that.
My attempt was as follows:
If $$|f'(\overline z)|^k<1$$, then there would exist a neighbourhood of $$\overline{z},\, U\subset V$$, where $$f^k(z)\rightarrow \overline{z},\,z\in U$$. But then by analicity, $$f^k(z)\rightarrow \overline{z}$$ in all compact subsets of $$V$$ containing $$U$$. But by hypothesis $$f^k$$ isn't normal in $$V$$ (since it is a negihbourhood of $$w$$), so that $$f^k$$ may not converge in every compact subset of $$V$$,...., but this argument doesn't include the compact subsets of $$V$$ that don't contain $$U$$... Nor does it work if $$|f'(\overline z)|^k=1$$...
Any ideias?

• what's $\exists \not = w$? – mathworker21 Sep 21 '18 at 17:58
• @mathworker21 My bad, fixed it – MathNewbie Sep 21 '18 at 18:22