I am not sure about all the details, but since you assume to deal with complex analytic sets whose codimension is at least 2, your result is similar to Hassler Whitney's version ([1] ยง2.5, pp. 50-51, theorem 5B) of Hartogs' theorem on separate analyticity:
Theorem. Let $H\subset \Bbb C^n$ be a open set and let $V$ be an analytic set of codimension $\ge 2$ in $H$. Then any holomorphic function $f$ in $H\setminus V$ has a unique holomorphic extension to the whole $H$.
His proof does not uses directly Cauchy's formula, but he uses other results he proved earlier by using it. This perhaps is not a direct answer to your question, but it may help yourself checking it.
References
[1] Hassler Whitney, Complex analytic varieties, (English)
Addison-Wesley Series in Mathematics. Reading, Massachussets-Menlo Park, California-London-Don Mills, Ontario: Addison-Wesley Publishing Company. XII, 399 p. (1972), MR0387634, Zbl 0265.32008.
You forgot to assume that $f$ is not identically zero. Here is one way to argue:
Step 1. I will start with some general topology observations. Let $X$ be a topological space and $A\subset X$ a subset. One says that $A$ does not locally separate $X$ if for every $x\in X$ there is a neighbourhood $W_x$ of $x$ in $X$ such that $W_x\setminus A$ is connected.
I will now assume, mostly for convenience, that $X$ is metrizable. (In the case of interest, $X$ will be an open subset of ${\mathbb C}^n$.)
Lemma. Suppose that $X$ is connected, $A\subset X$ is closed, $int(A)=\emptyset$ and $A$ does not locally separate $X$. Then $X\setminus A$ is connected.
Proof. Let $V$ be a connected component of $X\setminus A$. I claim that $cl_X(V)=X$. If not, then, by connectedness of $X$, there is a boundary point $x$ of $cl_X(V)$ in $X$, and, hence, a sequence
$x_i\in X\setminus cl(V)$ converging to $x$. Since $A$ is closed and has empty interior in $X$, the sequence $x_i$ can be chosen so that $x_i\notin A$ for each $i$. Let $W_x$ denote a neighborhood of $x$ such that
$W_x\setminus A$ is connected. Since $x$ is in the closure of $V$, the intersection $W_x\cap V\subset W_x\setminus A$ is also nonempty. Furthermore, for some $i$, $x_i$ is in $W_x\setminus A$. By the connectedness of $W_x\setminus A$, it follows that $x_i$ lies in the connected component $V$ of $X\setminus A$. A contradiction.
Thus, $cl_X(V)=X$. Since $cl_X(V)\subset V \cup A$ (as $V$ is a connected component of $X\setminus A$), it follows that $X=V\cup A$, i.e. $X\setminus A$ is connected. qed
I will now turn to your question.
Let $X$ be an open connected subset of ${\mathbb C}^n$, $f: X\to {\mathbb C}$ a nonconstant holomorphic function, $A:= f^{-1}(0)$. I will be using the fact that $A$ has empty interior in $X$ (already proven).
Step 2. Now, prove that $A$ does not locally separate $X$. Namely, given $x\in X$, take $W_x$ to be an open ball centered at $x$ and contained in $X$. Verify connectedness of $W_x\setminus A$ by looking at the intersections of complex lines $L$ with $W_x$: For each pair of distinct points $p, q$ in $W_x\setminus A$ take the complex line $L$ containing $p, q$. Then $f|_{L\cap X}$ is nonconstant, i.e. $L\cap A$ is a discrete subset of $L\cap X$. (I will leave it to you as an exercise.) From this, conclude path-connectedness of $L\cap (W_x\setminus A)$ and, hence, path-connectedness of $W_x\setminus A$. (Again, an exercise.)
- Now, combine Lemma and and Step 2 to conclude that the zero-level set of a nonconstant holomorphic function does not separate the domain, as required.
Best Answer
The existence of such a $\delta$ follows from the fact that $e_j' < e_j$. If we had equality then we might not be able to choose the $\delta$ parameter.
Edited: As you say, the choice of $\delta$ guarantees that the set lies in $B_{\epsilon}(0) \setminus B_{\epsilon '}(0)$. The reason we choose this set to restrict only the first two coordinates is that these coordinates are all we need to establish the holomorphicity of the Laurent series expansion of $f$ in the entire $B_{\epsilon}(0)$. We use $z_1$ to write the Laurent expansion as a polynomial in $z_1$ with holomorphic coefficients $\alpha_i(w)$. The restriction on $z_2$ and the preceding lemma allows us to analyze the coefficients $\alpha_i(w)$ and conclude that they must vanish identically on $B_{\epsilon}(0)$ for $i < 0$. Thus the Laurent expansion for $f$ actually yields a function which is holomorphic on all of $B_{\epsilon}(0)$ which agrees with $f$ on $B_{\epsilon}(0) \setminus B_{\epsilon'}(0)$.
As for a domain in $\mathbb{C}$ for which this fails, the classic example is the unit disk minus the origin $D =\mathbb{D} - \{0\}$ and the function $f(z) = \frac{1}{z}$; $f$ cannot be extended to the origin but is holomorphic in $D$.