I'm trying to prove Lemma 10.4 of Milnor's Dynamics in One Complex Variable, Third Edition, using a power series argument rather than the geometric argument proposed in the text.
Let $f$ be a holomorphic function $D \in \mathbb{C} \rightarrow \mathbb{C}$ fixing $0 \in D$. Denote by $\lambda := f'(0)$ the derivative of $f$ at $0$ and let it be a primitive $q$-th root of unity, $\lambda^q = 1$.
Consider the $q$-fold iterate of $f$, $f^{\circ q} := f \circ f \circ \dots \circ f$ $q$ times, and consider the case it's not the identity map. In this situation $0$ is called parabolic fixed point of $f$.
Clearly $f^{\circ q}$ fixes $0$, and by chain rule its derivative at $0$ is $1$: $ \left(f^{\circ q} \right)'(0) = \left(f'(0) \right)^q = \lambda^q = 1$. The origin is hence a parabolic fixed point also for the $q$-fold iterate.
Expand the (holomorphic) $q$-fold iterate around the origin: $$ \begin{split} f^{\circ q}(z) &= 0 + 1 \, z + a z^{n+1} + \text{higher terms} \\ & = z+a z^{n+1} + \text{h.t.} \\ & = z \, (1+az^n) + \text{h.t.} \end{split} $$ for some $a \neq 0 \in \mathbb{C}$ and integer $n\geq 1$. Per definition the integer $n+1$ is the multiplicity of $0$ as a fixed point of $f^{\circ q}$.
Lemma The integer $n$ is a multiple of $q$.1
The proof in the text uses a geometric argument making use of previous results 2, but the lemma can also be proved by a direct power series computation observing that $f \circ f^{\circ q} = f^{\circ q} \circ f $ - or so I have been told.
All I could think of is to expand $f$ around $0$ as $f(z) = \lambda z + \dots$, but also using the hint this doesn't seem to lead me anywhere.
Hints?
1 Or, as phrased in the text, the multiplicity $(n+1)$ is congruent to $1$ modulo $q$, which indeed means that $n+1-1$ is a multiple of $q$.
2 The integer $n$ corresponds to the number of attracting vectors of the iterate function at the fixed point. It can be shown that multiplication by $\lambda = e^{\frac{2\pi i j}{q}}$ for some $j = 0, \dots, q-1$ permutes these vectors, whose number must hence be a multiple of $q$.
Expand around the fixed point and compare the coefficients of $f \circ f^{\circ q}$ and $f^{\circ q} \circ f$. Let $f^{\circ q} (z) = z + a z ^{n+1} + \dots$ and $f(z) = \lambda z + \sum_{m=2}^{n+1}b_m z^m + \dots$, where $a$ and $b_m$ are the standars Taylor coefficients. Up to order $n+1$ one has
$$f \circ f^{\circ q} = \lambda z + \sum_{m=2}^{n}b_m z^m + (b_{n+1}+ a \, \lambda \, ) \, z^{n+1} + \dots$$
$$f^{\circ q} \circ f = \lambda z + \sum_{m=2}^{n}b_m z^m + (b_{n+1}+a \, \lambda^{n+1} \, ) \, z^{n+1} + \dots$$
Since $f \circ f^{\circ q} = f^{\circ q} \circ f$ coefficients must agree at all orders, so $\lambda = \lambda^{n+1}$ i.e. $\lambda^n = 1$. Since $\lambda$ is a primitive $q$-th root of unity, $n$ must be a multiple of $q$. $\quad \square$