complex analysis - Proof of identity principle for power series

Former: I want to proof that a convergent power series in $B_r(0)$ has only coefficents that are equal to zero (identity priniciple for power series).
Let's write the power series as $P(z) = \sum_{n=0}^{\infty}a_n z^{n}$.
I read in some book that I can write the coefficents like that:

$$\begin{align*} a_n = \frac{P^{(n)}(0) }{n!} \, , \end{align*}$$
but I'm not sure why. Anyway then I get
$$\begin{align*} P(z) = \sum_{n=0}^{\infty}\frac{P^{(n)}(0) }{n!} z^{n} = \exp(z) \sum_{n=0}^{\infty}P^{(n)}(0) \overset{?}= 0 \end{align*}$$
Maybe I'm on the wrong way. Can somebody give me a hint how to start?

Edit: I want to proof that a convergent power series that is constant zero can only have zero coefficents. Still I can write: $P(z) = \sum_{n=0}^{\infty}a_n z^{n} \equiv 0$ and now I have

$$\begin{align*} 0 \equiv \sum_{n=0}^{\infty}a_n z^{n} \end{align*}$$

Answer

The following result is standard: suppose that $P(z)$ converges for some $z_0 \neq 0$. Let $r = |z_0|$. Then $P(z)$ converges for all $z$ in the open ball $B$ centered at $0$ with radius $r$, uniformly on compact sets. In particular, $P(z)$ is a continuous function on $B$.

Now, suppose not all the coefficients $a_n$ are zero. Choose $a_n \neq 0$ with $n$ minimal. Then

$$P(z) = z^n(a_n + a_{n+1}z + \cdots)$$

for all $z \in B$. For the same reason as above, $g(z) = a_n + a_{n+1}z + \cdots$ is continuous on $B$, with $g(0) \neq 0$. So there is a punctured neighborhood of $0$ in which $g(z)$, and hence $P(z)$, is not zero.