Hi Dmitrii, though I didnt completely follow your proof it triggered coming up with a much more simplified proof. As a prerequisite we only need Picard's big theorem:
Picard's big theorem applied to entire functions. Each entire non-polynomial function takes on every complex number with at most one exception infinitely often.
First it can be shown that \( J \) (as an entire function) does not even omit one value from \( \mathbb{C} \). (I give the proof in a next post, though this assertion is not really necessary for the following conclusions.)
We know that \( J(k)=k \) for each \( k\in\mathbb{Z} \), but by the above theorem for each \( k\in\mathbb{Z} \) there have to be infinitely other \( z\in \mathbb{C}\setminus \mathbb{Z} \) with \( J(z)=k \).
Hence \( J(\mathbb{C}\setminus\mathbb{Z})=\mathbb{C} \).
That means if we have a function \( g(z)=f(J(z)) \), where \( f \) is a superexponential with singularities only at \( \{z\in\mathbb{Z}:z\le -2\} \), then \( g \) has singularities outside \( \{z\in\mathbb{Z}:z\le -2\} \).
Picard's big theorem applied to entire functions. Each entire non-polynomial function takes on every complex number with at most one exception infinitely often.
First it can be shown that \( J \) (as an entire function) does not even omit one value from \( \mathbb{C} \). (I give the proof in a next post, though this assertion is not really necessary for the following conclusions.)
We know that \( J(k)=k \) for each \( k\in\mathbb{Z} \), but by the above theorem for each \( k\in\mathbb{Z} \) there have to be infinitely other \( z\in \mathbb{C}\setminus \mathbb{Z} \) with \( J(z)=k \).
Hence \( J(\mathbb{C}\setminus\mathbb{Z})=\mathbb{C} \).
That means if we have a function \( g(z)=f(J(z)) \), where \( f \) is a superexponential with singularities only at \( \{z\in\mathbb{Z}:z\le -2\} \), then \( g \) has singularities outside \( \{z\in\mathbb{Z}:z\le -2\} \).