Schanuel's Conjecture Implies Transcendence of Log Pi

Theorem
Let Schanuel's Conjecture be true.

Then $\ln \pi$ is transcendental.

Proof
Assume the truth of Schanuel's Conjecture.

Let $z_1 = \ln \pi$, $z_2 = i \pi$.

Since $z_1$ is wholly real and $z_2$ is wholly imaginary, they 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 $\Q \left({\ln \pi, i \pi, \pi, e^{i \pi}}\right)$ has transcendence degree at least $2$ over $\Q$.

However, by Euler's Identity, $e^{i \pi} = -1$ is algebraic.

Also, $i \pi$ and $\pi$ are not algebraically independent, as they satisfy $x^2 + y^2 = 0$, where $x = i \pi$ and $y = \pi$.

Therefore, if Schanuel's Conjecture holds, $\ln \pi$ must be transcendental.