As the title says, is $2^e$ in the field $\mathbb{Q}(e)$? I mostly study analysis, but this came up trying to answer someone else's question.
So far, my idea has been to suppose it's true and use the fact that $\mathbb{Q}(e)$ is isomorphic to the rational functions on $\mathbb{Q}$. Then, we have $2^e$ being some rational polynomial evaluated at $e$. I got stuck trying to evaluate at new places. Clearly we can do it at $m e$ for some integer $m,$ but that doesn't help much. I'd love to take derivatives or evaluate growth somehow, as those are easy ways to show $2^x$ is not a polynomial.
I also tried to make use of the fact that $e$ isn't doing anything interesting here, other than being transcendental. Again, nothing.
I've looked into theorems about when numbers are transcendental, but most (all?) of the theorems prove things are transcendental over $\mathbb{Q}$. This situation is different, and I don't see how to reduce it to the simpler case over $\mathbb{Q}$