Let $(X, \mathcal{O}_X)$ be a complex analytic space in the sense of Grauert, i.e., a $\mathbb{C}$-analytic ringed space which is locally isomorphic to a local model. We may assume that $X$ is a complex manifold for our purposes here.
Let $\mathcal{O}_X$ denote the sheaf of analytic functions on $X$ and $\mathcal{O}_X^{\ast}$ denote the associated multiplicative sheaf. For each $x \in X$, we define a morphism on stalks $\exp : \mathcal{O}_{X,x} \longrightarrow \mathcal{O}_{X,x}^{\ast}$ which assigns to each analytic function $f \in \mathcal{O}_{X,x}$ the non-zero analytic function $\exp(f)\in \mathcal{O}_{X,x}^{\ast}$.
Claim: $\ker(\exp) = 2\pi i \mathbb{Z}$, where $\mathbb{Z}$ is the stalk of the constant sheaf at $x$. The subscript $x$ is omitted for obvious reasons.
It is obvious that $2\pi i \mathbb{Z} \subset \ker \exp$,but it is not immediate that the reverse inclusion is also true. A standard argument, which can be found in Kaup and Kaup's text on several complex variables is of the following form:
For $f \in \ker(\exp)$, assume that $f(x) =0$. Expanding $\exp(z)$ into a power series centered at $x$, we obtained a stalk $g_x \in \mathcal{O}_{X,x}$ such that $\exp(f) = 1 +f - g f^2$. Since $\exp(f) =1$, it follows by induction that $$f = f^2 g = \cdots = f^{j+1} g^j \in \bigcap_{j=1}^{\infty} \mathfrak{m}_x^j = (0),$$ where the last line follows from elementary properties of analytic algebras and $\mathfrak{m}_x$ denotes the maximal ideal of the local ring $\mathcal{O}_{X,x}$. Explicitly, $\mathfrak{m}_x = \{ \varphi \in \mathcal{O}_{X,x} : f(x) =0 \}$.
I am curious as to whether anyone is aware of other ways of proving this result, perhaps in a more analytic or geometric manner?
