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 $\frac{d}{d(\bar{z})} Log(z)=0$. 

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.
