Every holomorphic function $D \rightarrow \mathbb C$ has a power series expansion valid on D Every holomorphic function $D \rightarrow \mathbb C$ has a power series expansion valid on D.
Using Cauchy Integral formula I get:
$$f^{n}(a)=\frac{n!}{2\pi i}\oint  \frac{f(z)}{(z-a)^{n+1}}dz$$
Thus the power series is:
$$f(z)=\sum f^{n}(a)(z-a)^n/n!$$ Subbing in Cauchy Integral:
$$f(z)=\sum \frac{n!}{2\pi i}\oint  \frac{f(z)}{(z-a)^{n+1}}dz \frac{(z-a)^n}{n!}$$
At this point I can interchange the signs, giving me:
$$\oint \sum \frac{n!}{2\pi i}\frac{f(z)}{(z-a)^{n+1}} \frac{(z-a)^n}{n!}dz$$
$$=\oint \sum_{n=0}^\infty\frac{f(z)}{(z-a)2\pi i}dz=\sum_{n=0}^\infty \oint \frac{f(z)}{(z-a)2\pi i}dz$$
$$=\sum_{n=0}^\infty f(a)$$
Can somebody help me out?
 A: From Cauchy's Integral Theorem, we have
$$f^{n}(z_0)=\frac{n!}{2\pi i}\oint_{\gamma}  \frac{f(z')}{(z'-z_0)^{n+1}}\,dz'$$
where $f(z)$ is analytic in and on the closed circle $\gamma$ with center $z_0$.
Then, the power series $\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$ can be written
$$\begin{align}
\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n&=\sum_{n=0}^\infty\left(\frac{1}{2\pi i}\oint_{\gamma}  \frac{f(z')}{(z'-z_0)^{n+1}}\,dz'(z-z_0)^n \right)\\\\
&\frac{1}{2\pi i}\oint_{\gamma} \frac{f(z')}{z'-z_0}\sum_{n=0}^\infty\left(\frac{z-z_0}{z'-z_0}\right)^n\,dz'\\\\
&=\frac{1}{2\pi i}\oint_{\gamma} \frac{f(z')}{z'-z_0}\frac{z'-z_0}{z'-z}\,dz'\\\\
&=\frac{1}{2\pi i}\oint_{\gamma} \frac{f(z')}{z'-z}\,dz'\\\\
&=f(z)
\end{align}$$ 
And we are done!
Note that we exploited the fact that $|z-z_0|<|z'-z_0|$ since over $\gamma$, $|z'-z_0|$ is the radius of the circle with center $z_0$ and $z$ is a point inside this circle.
A: I will suppose that $D$ is a nonempty open subset of $\Bbb C$. Cauchy's Integral Formula, as I know it, states that:

If $\Omega \subset \Bbb C$ is an open convex set, and $\gamma$ is a closed path in $\Omega$, and if $f \in H(\Omega)$, then:
$$f(z) \ \text{ind}_{\gamma}(z) = \frac1{2\pi i}\oint_{\gamma} \frac{f(\xi)}{\xi - z} d\xi$$

Let us show that $f$ has a power series expansion in the neighbourhood of any $a \in D$.
Let $a \in D$, then there exists $R > 0$ such that $D(a,R) \subset D$. Let $0 < r < R$, and let $\gamma$ be circle of center $a$ and radius $r$; i.e. $\gamma(t) = a + re^{it}$. We have that ind$_{\gamma}(z) = 1$ for any $z \in D(a,r)$, hence by applying Cauchy's Integral Formula (with $\Omega = D(a,r)$, which is convex and open), we have for all $z \in D(a,r)$,
$$f(z) = \frac{1}{2\pi i} \oint_{\gamma} \frac{f(\xi)}{\xi - z} d\xi$$
Let $\gamma^*$ denote Im$(\gamma)$ (i.e. $\{\gamma(t), t \in [0,2\pi]\}$). Fix $z \in D(a,r)$ and consider the sequence of functions $(u_n)$ defined by:
$$u_n(\xi) = \frac{f(\xi)}{\xi - a} \left( \frac{z - a}{\xi - a} \right)^n$$
As $f$ is continuous on $\gamma^*$ (which is compact), there is some $M > 0$ such that $|f(\xi)| \le M$ for all $\xi \in \gamma^*$.
Thus,
$$|u_n(\xi)| \le \frac{M}{r}\left( \frac{|z-a|}{r} \right)^n :=a_n$$
$(a_n)$ is nonnegative and $\sum_{n=0}^{\infty} a_n < \infty$ since $|z - a| < r$. Thus, by the Weierstrass M-test, $\sum_{n \ge 0} u_n$ is uniformly convergent over $\gamma^*$.
Now, we have:
$$\sum_{n=0}^{\infty} u_n(\xi) = \frac{f(\xi)}{\xi - z}$$
By uniform convergence over $\gamma^*$,
$$f(z) = \frac{1}{2\pi i} \oint_{\gamma} \frac{f(\xi)}{\xi - z} d\xi = \frac{1}{2 \pi i} \oint_{\gamma} \left( \sum_{n=0}^{\infty} u_n(\xi) \right) d\xi = \sum_{n=0}^{\infty} \underbrace{\frac1{2 \pi i} \left( \oint_{\gamma} \frac{f(\xi)}{(\xi - a)^{n+1}} d\xi \right) }_{:=c_n} (z-a)^n$$
Since $z$ was arbitrary inside $D(a,r)$, the representation $f(z) = \sum_{n=0}^{\infty}c_n (z-a)^n$ is valid for all such $z$. And since $r$ was arbitrary, this is also true for all $z \in D(a,R)$.
Consequently, $f$ has a power series expansion near any point in $D$.
