Prove that $\displaystyle (\phi^n )' (\alpha) = \prod_{ i=0}^{n-1}\phi'( \phi^i (\alpha)) $ .
In particular, $\alpha$ is a critical point of $\phi^n$ if and only if one of the points $\alpha, \phi(\alpha), \ldots , \phi^{n−1} (\alpha)$ is a critical point of $\phi$.
$\phi(z) \in \Bbb C(z)$ is a rational function and $\phi'(z)$ is the derivative of $\phi(z)$ Here $\phi^n=\phi\circ \ldots \circ \phi\ \text{($n$ times)}$
I was thinking that whether I have to use Taylor's theorem or not to prove the highlighted line or not.
As if we take the highlighted line to be true then:
$α$ is a critical point of $\phi^n$ iff $\displaystyle (\phi^n )' (\alpha)=0=\prod_{ i=0}^{n-1}\phi'( \phi^i (\alpha))$ iff $ \phi'( \phi^i (\alpha))=0$ for some $i$ iff one of $\alpha, \phi(\alpha), \ldots , \phi^{n−1} (\alpha)$ is a critical point of $\phi$.
Can you help me in proving the highlighted line?