Proof, [Lemma I.4.4, Zhu93]. By (c) and Proposition 27.6.5, there exists $g \in H(\complex; \complex)$ such that $f = e^{g}$. Since $f(0) = 1$, $g(0) = 0$. As $f'(0) = g'(0)f(0) = 0$, $g'(0) = 0$ as well.

From (c), $|g(z)| \le e^{|z|}$ for each $z \in \complex$. Thus for every $r > 0$ and $z \in B_{\complex}(0, r)$, $|g(z)| \le |g(z) - 2r|$, and

extends to a holomorphic function on $B_{\complex}(0, r)$ by Proposition 27.2.3. Since $|g_{r}(z)| \le 1$ for all $z \in \partial B_{\complex}(0, r)$, $g_{r}(z) \le 1$ for all $z \in B_{\complex}(0, r)$ by the Maximum Modulus Theorem.

Finally, since $|g_{r}(z)| \le 1$ for all $z \in \complex$ and $r > |z|$, $g = 0$ and $f = 1$.$\square$