Does there exist a polynomial sequence $P_n$ satisfying this? Rudin - RCA p.276 exercise 3

Does there exist a sequence of polynomials $P_n$ such that $P_n(0)=1$ for $n=1,2,...$, but $P_n(z)\rightarrow 0$ for every $z\neq 0$, as $n\to \infty$?

The chapter this exercise is included is about Runge's Theorem. However, Runge's Theorem gives no information about pointwise convergence, so I think it is true. But I have a trouble constructing one. How do I prove this?
 A: For $0 < r < R < +\infty$ and $0 < \varepsilon < +\infty$, let
$$A(r,R,\varepsilon) := \{ \rho \cdot e^{i\varphi} : r \leqslant \rho \leqslant R, \varepsilon \leqslant \varphi \leqslant 2\pi - \varepsilon\}.$$
For $n \in \mathbb{N}\setminus \{0,1\}$, let
$$K_n = \{0\} \cup e^{i/n}\cdot A\bigl(\tfrac{1}{n},n,\tfrac{1}{n+1}\bigr).$$
$\mathbb{C}\setminus K_n$ is connected for all $n$, and the function
$$f_n \colon z \mapsto \begin{cases}1 &, z = 0 \\ 0 &, z \in K_n\setminus\{0\} \end{cases}$$
is holomorphic on $K_n$ (i.e., it's the restriction to $K_n$ of a holomorphic function defined on some open neighbourhood of $K_n$), so by Runge's theorem there is a polynomial $Q_n$ such that for all $z \in K_n$ we have $\lvert Q_n(z) - f_n(z)\rvert \leqslant 2^{-n}$. Let $P_n(z) = Q_n(z)/Q_n(0)$, then $P_n(0) = 1$ and $\lvert P_n(z) - f_n(z)\rvert \leqslant 2^{1-n}$ for all $z\in K_n$.
Since every $z \in \mathbb{C}\setminus 0$ belongs to $K_n$ for all large enough $n$, it follows that the sequence $(P_n)$ has the desired properties.
