Let $U$ be an open subset of the complex plane $\mathbb{C}$. Denote by $H(U)$ the $\mathbb{C}$-algebra of holomorphic functions $U\to\mathbb{C}$. The first Weyl algebra $A_1 = A_1(\mathbb{C})$ acts on $H(U)$ in the obvious way: the polynomials acts by multiplication and the derivation $\partial$ acts as the differentiation operator: $\partial f = f'$. Thus $H(U)$ is a left $A_1$-module.
I need to show that $H(U)$ is not a cyclic $A_1$-module. For this, assume the opposite, and let $h\in H(U)$ be such that $H(U)=A_1 h$. It is easy to see that $h$ cannot be a constant function (otherwise $H(U) = \mathbb{C}[x]$, which is absurd since $\exp\in H(U)$). I will be done if I can show that $\exp\circ h\not\in A_1 h$, that is, if $h$ does not satisfy a differential equation of the form $$ e^{h(z)} = \sum_{j=0}^n f_j(z)\frac{d^j h}{dz^j}(z), \quad z\in U, $$ where $f_j\in \mathbb{C}[x]$. But I'm stuck here. I have really no idea about how to prove this last assertion.
This proof is suggested by Coutinho in Exercise 4.7 on Chapter 5 in his book A primer of Algebraic $D$-modules.
Any help will be appreciated!
A cyclic module over the Weyl algebra is in particular countable-dimensional, and $H(U)$ is uncountable-dimensional, e.g. the functions $\exp(rz)$ for $r \in \mathbb{C}$ are linearly independent (exercise; there are several nice ways to do this, and the same result holds for the functions $\exp(rx), r \in \mathbb{R}$ on any open interval in $\mathbb{R}$ also, so the continuous, smooth, or analytic functions on an open interval are also uncountable-dimensional, and the functions $\exp(-rx^2), r > 0$ show the same for the Schwartz functions).