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*} 
 A: 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.
A: This factorization is nonsense:
$$
\sum_{n=0}^{\infty}\frac{P^{(n)}(0) }{n!} z^{n} = \exp(z) \sum_{n=0}^{\infty}P^{(n)}(0)
$$
It is an elaborate version of this false factorization:
$$
(a_1b_1+a_2b_2) = (a_1+a_2)\;(b_1+b_2)
$$
A: The proof provided by D_S is correct and in many ways is more insightful than the conventional proof I am offering here. But I am still offering it as it has its own insight.
If $P(z)$ converges in a neighborhood of $0$, then it is differentiable and the derivative is $P'(z) = \sum_{n=1}^{\infty}na_nz^{n-1}$, which has the same radius of convergence and therefore is also differentiable, etc. In general,
$$P^{(k)}(z) = \sum_{n=k}^{\infty}\frac{n!}{(n-k)!}a_nz^{n-k} = \sum_{m=0}^{\infty}\frac{(m+k)!}{m!}a_{n+k}z^m$$
So $P^{(k)}(0) = k!a_k$, or $a_k = \frac{P^{(k)}(0)}{k!}$.
If $P(z) = Q(z)$ in some neighborhood of $0$, then they have the same derivatives of all orders at $0$. Therefore
$$b_k = \frac{Q^{(k)}(0)}{k!} = \frac{P^{(k)}(0)}{k!} = a_k$$
for all $k$.
A: Do you mean this statement? 
" Every convergent power series has vanishing coefficients only. "
This is false. If it is true, all analytic mappings should vanish identically on each convergence domain.
However, exp is an entire function and this never vanish
on the complex plane.
