Holomorphic implies analytic I am trying to understand the proof that a holomorphic function is analytic but I have some questions. 
The claim is the following: Let $D\subset \mathbb C$ be connected and open and let $f: D \to \mathbb C$ be holomorphic. Then $f$ is analytic.
The proof (what I understand so far):
Let $z_0 \in D$. We must find an open ball around $z_0$ contained in $D$ and such that $f$ equals its Taylor series on the open ball.
Step 1: Pick any $\varepsilon > 0$ such that $B(z_0, \varepsilon) \subset D$ and apply Cauchy's integral formula to get 
$$ f(z_0) = {1\over 2 \pi i} \int_{\partial B(z_0, \varepsilon)} {f(z) \over z_0 - z} dz $$ 
Step 2: Expand the geometric series ${1\over 1 - {z \over z_0}}$ to get 
$$ f(z_0) = {1\over 2 \pi i} \int_{\partial B(z_0, \varepsilon)} f(z) {1\over z_0}\sum_{n}\left ({z\over z_0}\right )^n dz $$ 
Step 3: Use that the series converges absolutely (because it converges and so converges absolutely for smaller radii by Abel's theorem?) so that $\int$ and $\sum$ commute:
$$ f(z_0) = {1\over 2 \pi i} \sum_{n}z^n\int_{\partial B(z_0, \varepsilon)} f(z) {1 \over z_0^{n+1}} dz = \sum_{n} c_n z^n $$ 
This proves that if $f$ is holomorphic then it is infinitely differentiable. 

But how can I show from here that $f^{(n)}(z_0) = {n! \over 2 \pi
> i}\int {f(z) \over (z- z_0)^{n+1}}dz$?

My idea was to apply the identity theorem to the Taylor series of $f$ and this power series above but the problem is that at this point I don't know that the Taylor series converges to $f$. And another problem is that the proof of the identity theorem might rely on the theorem that holomorphic implies analytic. 
Also:

Is every step above 100% correct?

 A: You have the wrong expansion of the series (which you can see by the fact that you're worried how the expansion works: $z$ and $z_0$ appear unrelated).
Let's start again. WLOG, $z_0=0$, because other points are just a translation away, and it makes the algebra simpler. Let $\lvert w \rvert=\varepsilon>\lvert z \rvert$. The Cauchy integral formula says
$$ f(z) = \frac{1}{2\pi i}\int_{\lvert w \rvert = \varepsilon} \frac{f(w)}{w-z} \, dw \tag{1} $$
(to get this the right way around, remember that the proof looks at the function $\frac{f(w)-f(z)}{w-z} \to f'(w)$ as $z \to w$), and since $\lvert z \rvert<\lvert w \rvert$, the following series expansion holds:
$$ \frac{1}{w-z} = \frac{1}{w}\frac{1}{1-(z/w)} = \frac{1}{w} \sum_{n=0}^{\infty} \left(\frac{z}{w} \right)^n. $$
Inserting this into the integral,
$$  f(z) = \frac{1}{2\pi i}\int_{\lvert w \rvert = \varepsilon} \sum_{n=0}^{\infty} z^n \frac{f(w)}{w^{n+1}} \, dw. $$
Now, to swap the sum and the integral, note that $f$ is holomorphic, so it is continuous on the circular integration contour and hence bounded above. Therefore the sum is bounded uniformly in the domain of integration by a constant multiple of a convergent geometric series, and the Weierstrass M-test says this means it converges uniformly absolutely, which is enough to allow us to swap the sum and the integral, which gives
$$ f(z) = \sum_{n=0}^{\infty} z^n c_n, \quad c_n = \frac{1}{2\pi i} \int_{\lvert w \rvert = \varepsilon} \frac{f(w)}{w^{n+1}} \, dw $$
Term-by-term differentiation (which is allowed inside the radius of convergence of a power series by some arcane result involving more uniform convergence) then tells us that $c_n=f^{(n)}(0)/n!$. Another way to get the coefficients is to differentiate (1) $n$ times with respect to $z$, using that the integrand is differentiable on the contour, which is easy to show directly. The above all applies for any $z$ with $\lvert z \rvert < \varepsilon$, so $f$ is equal to its power series in the open ball $B(0,\varepsilon)$.
