Is $ e^{e^q + q} $ a transcendental number for every rational number q? [closed]

Is $ e^{e^q + q} $ a transcendental number for every rational number q?

This question was also asked here: https://www.wetenschapsforum.nl/viewtopic.php?t=213111&start=30

Some extra background: I was wondering whether the following definitions lead to a mathematically interesting family of function classes.

(i) Let the function class $ \mathbb{D}^0 $ be the set of all continuous functions f from $ \mathbb{Q} $ to $ \mathbb{R} $ for which f(x) is transcendental for all rational x.

(ii) Let the function class $ \mathbb{D}^n $ for positive natural numbers n be the set of all continuous functions f from $ \mathbb{Q} $ to $ \mathbb{R} $ for which the n-th derivative of f exists and is an element of $ \mathbb{D}^0 $.

Now on another website (Wetenschapsforum.nl) somebody suggested $ f: \mathbb{Q} \to \ \mathbb{R} ; f(x) = e^{e^x} $ as a member of $ \mathbb{D}^1 $. And that suggestion lead to the question of this topic.

Solution 1:

Assuming Schanuel's conjecture:

The case $q=0$ is trivial, so assume $q \ne 0$, so $q$ and $e^q$ are linearly independent over $\Bbb Q$, so the conjecture says that $\Bbb Q(q, e^q, e^q, e^{e^q})$ has transcendence degree at least $2$.

But the field $\Bbb Q(q, e^q, e^q, e^{e^q})$ is the same as the field $\Bbb Q(e^q, e^{e^q+q})$, so $e^{e^q+q}$ must be transcendental.