Show that $f'(f(z))$ omits $0$ and $c$ (Entire functions)

Question

Assume that $f$ is entire and that $z\to f(f(z))$ has no fixed points. Consider $$g(z)=\frac{f(f(z))-z}{f(z)-z}$$ which is an entire function. I have shown that $g$ is constant i.e. $$f(f(z))=cf(z)+z(1-c)$$ for some $c\in C\setminus\{0,1\}$. Why does it follow that $f'(f(z))$ omits $0$ and $c$?

Answer 1

Differentiating the identity

$$f(f(z)) = cf(z) + z(1-c)$$

yields

$$f'(f(z))\cdot f'(z) = cf'(z) + (1-c)\,.\tag{$\ast$}$$

Since $1-c \neq 0$, it follows that $f'(f(z))$ never attains the value $c$. Therefore we can rearrange $(\ast)$ into

$$f'(z) = \frac{1-c}{f'(f(z)) - c}\tag{$\ast\ast$}$$

which shows that $f'$ never attains the value $0$.