I will assume we can write
$$f(z) = \frac{c}{z_0-z} + \sum_{n=0}^{\infty} b_n z^n$$
for some value of $c \ne 0$, and $\lim_{n \to \infty} b_n = 0$. Then
$$f(z) = \sum_{n=0}^{\infty} a_n z^n$$
where
$$a_n = b_n + \frac{c}{z_0^{n+1}}$$
Then
$$\begin{align}\lim_{n \to \infty} \frac{a_n}{a_{n+1}} &= \lim_{n \to \infty} \frac{\displaystyle b_n + \frac{c}{z_0^{n+1}}}{\displaystyle b_{n+1}+ \frac{c}{z_0^{n+2}}}\\ &= \lim_{n \to \infty} \frac{\displaystyle \frac{c}{z_0^{n+1}}}{\displaystyle \frac{c}{z_0^{n+2}}}\\ &= z_0\end{align}$$
as was to be shown. Note that the second step above is valid because $z_0$ is on the unit circle.
For a nonsimple pole, we may write
$$f(z) = \frac{c}{(z_0-z)^m} + \sum_{n=0}^{\infty} b_n z^n$$
for $m \in \mathbb{N}$. It might be known that
$$(1-w)^{-m} = \sum_{n=0}^{\infty} \binom{m-1+n}{m-1} w^n$$
Then
$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{n+1}{n+m} z_0$$
EDIT
@TCL observed that we can simply require that $b_n z_0^n$ goes to zero as $n \to \infty$. Then for a simple pole
$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{\displaystyle b_n z_0^n + \frac{c}{z_0}}{\displaystyle b_{n+1} z_0^n + \frac{c}{z_0^2}}$$
which you can see goes to $z_0$.
From the expression for $P(z)/z^n$ it follows that
$$
\lim\limits_{z\to\infty}\frac{P(z)}{z^n}=a_n\neq 0
$$
Therefore, by definition of limit, by choosing $|z|$ large enough, say $|z|>R$ we can make $\left|\frac{P(z)}{z^n}\right|$ as close to $|a_n|$ as we want, say between $|a_n/2|$ and $|3a_n/2|$, so
$$
\left|\frac{P(z)}{z^n}\right|\ge c:=\frac{a_n}{2},\quad |z|>R
$$
and, consequently, $|P(z)|\ge cR^n$ if $|z|>R$, so $P$ is bounded below by the positive constant $cR^n$ outside of the ball of radius $R$. Inside, $|P|$ is clearly bounded below, being a positive continuous function on a compact set.
Best Answer
His statement is a precise way of saying that the leading term of a polynomial dominates for large enough $|z|$. Assume $a_{n}\ne 0$. For the polynomial in question, whenever $0 < r \le |z|$, one has $1 \le |z|/r$ and $$ \begin{align} |a_{n-1}z^{n-1}+\cdots+a_{0}| & \le |a_{n-1}||z|^{n-1}+\cdots+|a_{0}| \\ & \le \left(\frac{|a_{n-1}|}{r} +\frac{|a_{n-2}|}{r^{2}} +\cdots+\frac{|a_{0}|}{r^{n}}\right)|z|^{n} \end{align} $$ The term in parentheses in the last expression tends to 0 as $r\rightarrow \infty$. So you can choose $R$ large enough that this term is bounded by $|a_{n}|/2$ for $r > R$. Then, for $r > R$, $$ |P(z)| \ge |a_{n}||z|^{n}-|a_{n-1}z^{n-1}+\cdots a_{0}| > |a_{n}||z|^{n}-\frac{|a_{n}|}{2}|z|^{n} \ge \frac{|a_{n}|}{2}|z|^{n},\;\; |z| > R. $$