Is $2^{\sqrt2}\cdot\pi$ known to be transcendental number?

Question

Specific Question

I know that $2^{\sqrt2}$ and $\pi$ are each transcendental. But is it known that their product or sum are also transcendental?

Exposition

Schanuel's conjecture is a strengthening of the Lindemann-Weierstrass-theorem. It states

If $\lambda_1,\cdots \lambda_n$ are complex numbers linearly independent over $\mathbb Q$, then $$\mathbb Q(\lambda_1,\cdots \lambda_n,e^{\lambda_1},\cdots ,e^{\lambda_n})$$ has transcendental degree at least $n$ over $\mathbb Q$

Some useful information would be:

1. Is Schanuel's conjecture sufficient to establish that the sum and product are transcendental?
2. If we don't assume Schanuel's conjecture, is this still open question?
3. If it is still open, is there any partial progress towards this result?

Answer 1

The best answer: We don't know [without taking unproven assumptions]. This number $\pi 2^\sqrt 2$ is indeed beyond the current state of the art.

It is the expectation that any arbitrary number should be transcendental and it should be the expectation that an arbitrary representation of a number doesn't have a proof of it's transcendence.

Your forms above are "Schanuel Conjecture" demonstrably transcendental. It is exactly as paste bee puts it in the comments.

This would follow from Schanuel's conjecture: $ \log 2, \sqrt{2} \log 2 $ and $ \frac{1}{2} \pi i $ are linearly independent over $ \mathbb{Q} $, and so $ \mathbb{Q}(\log 2, \sqrt{2} \log 2, \frac{1}{2} \pi i, 2, 2^{\sqrt{2}}, i) = \mathbb{Q}(\sqrt{2}, i, \log 2, \pi, 2^{\sqrt{2}}) $ has transcendence degree at least 3 over $ \mathbb{Q} $, meaning $ \log 2, \pi $ and $ 2^{\sqrt{2}} $ are algebraically independent.

I suspect that question 2 is unknown. I am not sure we have a demonstration that $2^\sqrt{2}\pi$ is irrational.