How do i prove that there exists a holomorphic function $f: \mathbb C \setminus\{0\} \rightarrow \mathbb C $ so that $f(\frac{1}{n}) = 0$ for all $n \in \mathbb N$ that is not the constant nullfunction.
Why is that not a contradiction to the Identity theorem?
It suffices to construct an example. For example, $$ f(z) = \sin\left(\frac{\pi}{z}\right) $$ This does not contradict the identity theorem because $f$ and the null-function agree only on $\{1/n:n \in \Bbb Z \setminus \{0\}\}$, which is not an open set (nor does it contain an open set). That is, $\{f = 0\}$ has an empty interior.