When I first learned about this result I was completely stunned that there is no holomorphic function $f(z)$ on $\Bbb{C}\setminus\{0\}$ such that $\exp f(z) = z$. What are some interesting proofs of this? Here are two I know of.
Proof 1: Assume we have such an $f(z)$. Then restricted to $\Bbb{C}\setminus (-\infty,0]$ we must have $f(z) = \operatorname{Log} z + 2\pi k i$ for some $k \in \Bbb{Z}$. The big log stands for principal log. But now the $\lim_{z \to a} f(z)$ for any $a \in (-\infty,0]$ does not exist (approaching from the top, the argument approaches $\pi$ but from the bottom it approaches $-\pi$).
Proof 2: If we had a holomorphic function $f(z)$ such that $\exp f(z) = z$ then differentiating gives $f'(z) = 1/z$ for all $z \neq 0$. Integrating on $|z|=1$, the left gives $0$ by Cauchy's Theorem but the right is $2\pi i$ which is a contradiction.
Chances are that all proofs of this will be morally equivalent. But here is an interesting way to phrase it. Consider the exponential sheaf sequence on $X=\mathbf C-\{0\}$ (viewed as a complex manifold):
$$0 \to \mathbf Z \to \mathcal O \xrightarrow{\text{exp}(2\pi i\cdot)} \mathcal O^\times \to 1.$$
The long exact sequence of cohomology begins with
$$0 \to \mathbf Z \to \mathcal O(X) \xrightarrow{\text{exp}(2\pi i\cdot)} \mathcal O^\times(X) \to H^1(X, \mathbf Z) \to \dots.$$
The connecting map $\mathcal O^\times(X) \to H^1(X, \mathbf Z) = \mathbf Z$ is given by $f \mapsto \frac{1}{2\pi i}\int_C df/f$, where $C$ is the unit circle oriented counter-clockwise. (By the residue theorem, this really is an integer.) Now, since $\frac{1}{2\pi i}\int_C dz/z = 1$, the function $z$ cannot be in the image of $\mathcal O(X) \xrightarrow{\text{exp}(2\pi i\cdot)} \mathcal O^\times(X)$.