Is the following function holomorphic? 
Let $f(z)$ be continuous on some connected open set $U \subseteq \mathbb{C}$. Suppose $e^{f(z)}$ is holomorphic. Does it follow that $f(z)$ is holomorphic? 

I personally do not see any counterexample, but I don't know how to go about proving this assertion if it is in fact true. 
The motivation for this question is to see if I can slickly prove that $Log(z)$ is holomorphic on its principal branch without a brute force calculation on the usual $Log(x+iy)$ formulation (which I can do; I'm just seeing if there's some other way). I see that $e^{Log(z)}=z$, so that $\frac{d}{d(\bar{z})}e^{f(z)}=0,$ but I don't know if I can therefore conclude $\frac{d}{d(\bar{z})} Log(z)=0$. 
 A: I think I can prove this when $f$ is differentiable (as a real function of two variables). Suppose $e^{f(z)}$ is holomorphic. The chain rule for the Wirtinger derivatives implies that
$$0 = \partial_{\bar{z}}(e^{f(z)}) = [(e^{z})_{z} \circ f]f_{\bar{z}} + \overbrace{[(e^{z})_{\bar{z}} \circ f]}^{0}\bar{f}_{\bar{z}} = e^{f(z)}f_{\bar{z}}$$
Since $e^{f(z)}$ never vanishes, we must have $f_{\bar{z}} = 0$, which implies that $f$ is holomorphic.
A: Under the additional condition that $U$ is simply-connected , I think the answer is yes:
Let $h(z)=e^{g(z)}$. Then $h(z)$ is holomorphic and non-zero in a simply-connected region,
so that $h(z)$ admits a holomorphic log $w(z):=log(h(z))$ , meaning that $e^{log(h(z))}=h(z)$. You can then show that $g(z)$
differs from $h(z)$ by a constant, so $g(z)$ is holomorphic. But I don't know how to 
weaken the condition that $U$ must be simply-connected .
I don't know if this is circular, but re your proof, since $f(z)=z$ is holomorphic,
away from $0$ , in a simply-connected region -- like the standard branch, $f(z)=z$ admits
a holomorphic log, and this is then $Log(z)$, up to a constant.
