Holomorphic function and open, closed set of zeroes $\{ z: f^n(z) = 0 , n = 1,2,3, \ldots \}$ and $f$ is holo on an open, connected region. $f^n$ is the ($n$-th) derivative of $f$ .
How do you show this is an open set by Taylor theorem?
It's closed as $x$ limit point implies for a given $n$ there exists $k(n)$ s.t. $k(n)$ is in the  region where
$f(x+k(n)) = 0$ and $|k(n)| < 1/n$.
Holomorphic functions are infinitely differentiable, and derivatives are continuous, therefore $f^n(x) =0$ as we can take the limit inside.
 A: Although the question is not very clear (as the comments show), this is perhaps what the OP intended to ask: 

Let $\Omega \subseteq \mathbb C$ be an open, connected region and let $f : \Omega \to \mathbb C$ be a holomorphic, nonzero function. Then there does not exist any point $z \in \Omega$ where the function and all its derivatives vanish. 

Here's a proof. Let $A$ be the set of all points $z \in \Omega$ such that $f^{(n)}(z) = 0$ for all $n \geqslant 0$. (I use $f^{(0)}$ to denote $f$.) We will show that $A$ is both open and closed, which implies (by connectedness of $\Omega$) that either $A = \Omega$ or $A$ is empty. Finally, if $f$ is nonzero, then $A$ cannot be $\Omega$; therefore, $A = \emptyset$. 
$A$ is closed. For any $n$, the set $B_n := \{ z \in \Omega \mid f^{(n)}(z) = 0 \}$ is closed since $f^{(n)}$ is continuous. Therefore, $A = \bigcap \limits_{n \geqslant 0} B_n$ is also closed. 
A is open. Suppose $z \in A$. Then since $f$ is analytic at $z$, there exists some $\varepsilon > 0$ such that for $w \in B(z, \varepsilon)$, we have
$$
f(w) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z)}{n!} \cdot (z-w)^n.
$$ 
But since $f^{(n)}(z) = 0$ by definition of $A$, $f$ is identically zero inside $B(z, \varepsilon)$. Therefore, all its derivatives also vanish inside the same ball; therefore, $B(z,\varepsilon) \subseteq A$. That is, $z$ is an interior point of $A$, and we are done.  
