Schanuel's Conjecture Implies Transcendence of Pi by Euler's Number

Theorem
Let Schanuel's Conjecture be true.

Then $\pi \times e$ is transcendental.

Proof
Assume the truth of Schanuel's Conjecture.

By Schanuel's Conjecture Implies Algebraic Indepdence of Pi and Euler's Number over the Rationals, $\pi$ and $e$ are algebraically independent over the rational numbers $\Q$.

That is, no non-trivial polynomials $f \left({x,y}\right)$ with rational coefficients satisfy:
 * $f \left({\pi, e}\right) = 0$

$\pi \times e$ is algebraic, then there would be a non-trivial polynomial $g \left({z}\right)$ with rational coefficients satisfying:
 * $g \left({\pi \times e}\right) = 0$

However, $f \left({x, y}\right) := g \left({x \times y}\right)$ would be a non-trivial polynomial with rational coefficients satisfying
 * $f \left({\pi, e}\right) = 0$

Which is contradictory to the earlier statement that no such polynomials exist.

Therefore, if Schanuel's Conjecture holds, $\pi \times e$ is transcendental.