Infinity as a super-attractive fixed point

Let $f(z)=p(z)/q(z)$ be a rational function such that $\deg(p) \geq \deg(q)+2$. Show that the point at $\infty$ is a super-attractive fixed point of $f$.

Here is my solution to the question. The only problem I had was at the last step where I had to say $\deg(q)\geq 1$.

Let $\deg(p)=n$ and $\deg(q)=m$. Given $n\geq m+2$, $\deg(f)>1\Rightarrow f(\infty)=\infty$ which means $\infty$ is a fixed point.

By the quotient rule,
$$f'(z)=\frac{p'(z)q(z)-p(z)q'(z)}{q(z)^2}\Rightarrow\deg\Big(f'(z)\Big)=\frac{m+n-1}{m^2}.$$

If we use the fact that $n\geq m+2\Rightarrow n-2\geq m\Rightarrow n-1>m$, we get

$$\deg\Big(f'(z)\Big)=\frac{m+n-1}{m^2}<\frac{(n-1)+n-1}{(n-1)^2}=\frac{2(n-1)}{(n-1)^2}\leq 1$$

when $n\geq 3$ which only happens when $m\geq 1$. Since

$$\deg\Big(f'(z)\Big)<1\Rightarrow f'(\infty)=0$$
which means infinity is a supper attractive fixed point.

@Person I don't quite get it. At some point, we've got to conjugate with $\varphi(z)=1/z$, don't we?

Here's a proof that $\infty$ is a super-attractive fixed point for a polynomial of degree 2 or greater. As we usually do when dealing with $\infty$ we conjugate with $\varphi(z)=1/z$ and examine the resulting function at zero. Specifically, let
$$f(z) = \sum_{k=0}^n a_k z^k$$
be a polynomial of degree $n\geq2$ and let $F(z)=1/f(1/z)$. Then,
$$
F(z) = 1/ \sum_{k=0}^n a_k (1/z)^k
= z^n/\sum_{k=0}^n a_k z^{n-k}.
$$
Then,
\begin{align}
F'(z) &= \frac{n z^{n-1}\sum_{k=0}^n a_k z^{n-k} - z^n \sum_{k=0}^{n-1} (n-k) a_k z^{n-k-1}}{\left(\sum_{k=0}^n a_k z^{n-k}\right)^2}.
\end{align}
Now, since the degree of $f$ is $n$, we know that $a_n\neq0$. Thus, we needn't worry about the denominator being zero when we plug in $z=0$ to get $F'(0)=0$.

Thank you that proof makes sense. I just have one question. I understand why conjugating with $\frac{1}{z}$ makes things easier but do we always have to conjugate when examining infinity?

@Person Yes, conjugating (typically with $\varphi(z)=1/z$ is exactly the way to deal with $\infty$. That turns questions questions dealing with $\infty$ into questions dealing with ordinary numbers.