Schanuel's Conjecture Implies Transcendence of Euler's Number to the power of Euler's Number

Theorem

Let Schanuel's Conjecture be true.

Then Euler's number $e$ to the power of itself:

$e^e$

is transcendental.

Proof

Assume the truth of Schanuel's Conjecture.

Let $z_1 = 1$, $z_2 = e$.

By Euler's Number is Irrational, $z_1$ and $z_2$ are linearly independent over $\Q$.

By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_2, e^{z_1}, e^{z_2}}\right)$ has transcendence degree at least $2$ over $\Q$.

That is, the extension field $\Q \left({1, e, e, e^e}\right)$ has transcendence degree at least $2$ over $\Q$.

However, $1$ is algebraic.

Therefore, if Schanuel's Conjecture holds, $e^e$ must be transcendental.

$\blacksquare$