Is a connected Reinhardt Domain which containg $0$ necessarely a polydisc? I'm studying several complex variables basics.
Roughly speaking: call $D\subseteq\Bbb C^n$ the set of points in which a given power series
$$
\sum_{\alpha\in\Bbb N^n}a_{\alpha}(z-z_0)^{\alpha}
$$
converges normally.
Call $B$ the set of $z$'s such that $|a_{\alpha}(z-z_0)^{\alpha}|<c$ for all $\alpha\in\Bbb N^n$.
We know that $D=\operatorname{int}B$.
This allow us to understand that $D$ must be a Reinhardt and complete domain (complete means that given $z\in D$ then, $w\in\Bbb C^n$ s.t. $|w_j|\le|z_j|\;\forall j=1,\dots,n\;\Longrightarrow w\in D$); furthermore these domains are characterized to be union of polydiscs centered in $0$ (easy to see).
We know moreover that holomorphic functions are exactly convergent power series on union of polydiscs.
In particular, over a Reinhardt complete domain, an holomorphic function is always represented as a sum of a power series centered in $0$.
Now my book states and proves a theorem which says that the last conclusion is nevertheless reached even if we don't ask the completeness.
Precisely the theorem says: 
Let $\Omega\subseteq\Bbb C^n$ a Reinhardt domain, connected, containing $0_{\Bbb C^n}$ and $f\in\mathcal{H}(\Omega)$. Then
$$
f(z)=\sum_{\alpha}\frac{f^{(\alpha)}(0)}{\alpha!}z^{\alpha}
$$
followed by a long proof.
My question is: it should follow directly from definition of Reinhardt domain that, an $\Omega$ like the one in the hypothesis of the theorem should be a polydisc. Hence, why can't we conclude immediately with this observation? (the book presents a nontrivial proof, technical and long).
 A: Everything is in the comments already, but I will spell it out in an answer here.
$$f(z,w) = \frac{1}{1-zw} = \sum_{0}^\infty z^kw^k$$ has domain of convergence $\{(z,w):|z||w|<1\}$, which is certainly not a polydisk, although it is connected and contains $0$.
It is true that if a connected Rienhardt domain $\Omega$ contains $0$, and $f$ is holomorphic on $\Omega$, then the power series of $f$ converges to $f$ normally on $\Omega$.  
From further discussion, it seems OP was confused about why the proof of this theorem is hard.  He thought "such a domain must be the union of polydisks centered at 0, and for polydisks it is easy".  
The key word missing in the theorem above, which makes it a hard theorem, is the word "complete" before Reinhardt.  For example, the theorem would even apply to the domain
$$\Omega = \{ (z,w) : |z|<1, |w|<1\} \cup \{(z,w): \frac{1}{2}<|z|<2, |w|<2\}$$
which is not a union of polydisks centered at $0$.  The fact that all holomorphic functions on $\Omega$ have their power series converge normally to them on all of $\Omega$ is really surprising.   You would think it would only converge on the largest polydisk contained in it, namely the polydisk of radius $1$.  The theorem guarentees that it converges on the whole thing.
In fact, every holomorphic function on a Reinhardt domain must converge on a (potentially) larger one:  namely the smallest complete log convex Reinhardt domain containing the given one.  This is an instance of the "Hartog's extension phenomenon".
