Complex Analysis Question from Stein The question is #$14$ from Chapter $2$ in Stein and Shakarchi's text Complex Analysis:

Suppose that $f$ is holomorphic in an open set containing the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $$\sum_{n=0}^\infty a_nz^n$$ denotes the power series expansion $f$ in the open unit disc, then $$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=z_0.$$

I've shown that we can take $z_0=1$ without a loss of generality, but I'm having trouble showing the proof otherwise. One of the problems I'm having is because we aren't told the definition of a pole except that it is a place where the function isn't holomorphic. Disregarding this fact, the other problem I'm running into is that I don't know the order of the pole.
Making some additional assumptions, including that the pole is simple so we can write $F(z)=(z-1)f(z)$ as a holomorphic function, we see that $$F(z)=-a_0+z(a_0-a_1)+z^2(a_1-a_2)+\cdots$$ This almost gets me to the end with these added assumptions, but I don't think it's quite enough (why do we know some of the $a_i$'s aren't $0$, for example).
On another note, if we know $\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ exists, then it is easy to see that $\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}=1=|z_0|$; I, however, do not see why the limit must exist.
Are there any hints that someone can provide? Even a solution would be nice, especially if one can avoid making any assumptions about what a pole is or is not.
EDIT: So there isn't any confusion, I know the definition of a pole and I'm inclined to believe that the problem, as stated, necessarily has a pole at $z_0$. The problem is that the exercise is in Chapter $2$, and poles are introduced in Chapter $3$.

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 $z_0$.
