an equivalent condition of entire function without zero

Question

Show that an entire function $f$ has no zeros if and only if there exists another entire function $g$ so that $f=e^g$.

For if part, easy because exp function has no zero and is of course entire. For only if part, I thought that f can be expanded as Taylor series centred at 0, and let g(z) = z*(f^(n)(0))^(1/n), so f = exp(g). Is that correct method? I think that g is multivalued. Is that a problem?

Answer 1

Hint: If $f=\exp(g)$, then $f'=g'f$. So define $\displaystyle g(z) = w_0+\int_0^z \frac{f'(w)}{f(w)} dw$, where $w_0$ is such that $f(0)=\exp(w_0)$, which exists since $f(0)\ne0$ and the image of $\exp$ is $\mathbb C \setminus \{ {0}\}$.