A complex number is said to be hypertranscendental if the one is not a zero of any entire function with all rational Maclaurin coefficients. Does there exist a transcendental number which is not hypertranscendental?

Let $z$ be an arbitrary complex number. Since $\mathbb{Q}[i]$ is dense in $\mathbb{C}$, we can choose a sequence of complex numbers $a_n$ such that $a_n< \frac{1}{n!}$ and $a_{n+1}  \frac{a_n}{z} \in \mathbb{Q}[i]$. Define an entire function $ f(t) = \sum_{n=0}^{\infty} a_n t^n $ Then the function $g(t) = (1t/z)f(t)$ is also entire, has coefficients in $\mathbb{Q}[i]$ and vanishes at $t = z$. Now let $h(t)$ be the function whose power series coefficients are the complex conjugates of the coefficients of $g(t)$. Then $g(t)h(t)$ is a power series with rational coefficients that vanishes at $z$. Hence there are no hypertranscendental numbers. 

