there exists $h$ holomorphic function in $U$ such that $f=e^h$? I am new to complex analysis, and I am unable to show the following result, if you have any tips please:
Let $f$ be a holomorphic function of a complex variable $z$ is  an open connected subset  $U$ such that $f(z)\neq 0$ for all $z\in U$.
I want to show that : there exists $h$ holomorphic function in $U$ such that  $f=e^h$
 A: 
Theorem. (Existence of Logarithm) If $f$ is a nowhere vanishing holomorphic function in a simply connected region $U$, then there exists a holomorphic function $F$ on $U$ such that $f(z) = e^{F(z)}$ for all $z\in U$.

Proof. Fix a point $z_0 \in U$. Define $$F(z):= \int_\gamma \frac{f'(w)}{f(w)}\, dw + c_0$$ where $\gamma$ is any curve in $U$ joining $z_0$ to $z$, and $c_0$ satisfies $e^{c_0} = f(z_0)$. By the homotopy version$\color{red}{^1}$ of Cauchy's theorem, $F$ is well-defined, i.e. it does not depend on the choice of $\gamma$. It is easy to check that
$$F'(z) = \frac{f'(z)}{f(z)}$$ for all $z\in U$. But then,
$$\frac{d}{dz}(fe^{-F})(z) = f(z) e^{-F(z)} (-F'(z)) + f'(z) e^{-F(z)} = 0$$
so that $f(z) = ce^{F(z)}$ for some constant $c\in \Bbb C$, for all $z\in U$. By the choice of $c_0$, we obtain $c = e^{c_0 - F(z_0)} = 1$, and hence
$$f(z) = e^{F(z)}$$ for all $z\in U$.

Footnotes:
$\color{red}{1.}$

Cauchy's Theorem: If $f$ is holomorphic in a simply connected region $\Omega\subset\mathbb C$, then $\int_\gamma f(z)\, dz  = 0$ for any closed curve $\gamma$ in $\Omega$.

A: To complete delta-divine's answer, one can quantify the problem as follows. Let $\subset \mathbb{C}$ be an arbitrary open region and $\mathcal{H}(U)$ be the set of holomorphic functions $U\to \mathbb{C}$ and $\mathcal{H}(U)^*$ that of functions vanishing nowhere (i.e. $f(U)\subset \mathbb{C}\setminus \{0\}$). One has
$$
e^{\mathcal{H}(U)}\subset \mathcal{H}(U)^*
$$
delta-divine's answer shows that this inclusion is equality if $U$ is simply connected.
For an example where the inclusion is strict, one has
$U=\mathbb{C}^*=\mathbb{C}\setminus \{0\}$ as $f(z)=z$ is not an exponential.
