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

Lemma
Let Schanuel's Conjecture be true.

Let $z_1 = \ln \ln \pi$, $z_2 = 1 + \ln \ln \pi$, $z_3 = \ln \pi$, $z_4 = e \ln \pi$, and $z_5 = i \pi$.

Then, $z_1$, $z_2$, $z_3$, $z_4$, and $z_5$ are linearly independent over the rational numbers $\Q$.

Proof
Assume the truth of Schanuel's Conjecture.

Now, we will prove that $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$.

Equivalently, they are linearly independent over the integers $\Z$.

Let $a, b \in \Z$ such that:
 * $a z_1 + b z_3 = 0$

Substituting:
 * $a \ln \ln \pi + b \ln \pi = 0$

Applying the exponential function to both sides:
 * $\left({\ln \pi}\right)^a \pi^b = 1$

By Schanuel's Conjecture Implies Algebraic Indepdence of Pi and Log of Pi over the Rationals, the above equation only has solution when $a = b = 0$.

Thus, $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$.

Since $z_5$ is wholly imaginary, $z_1$, $z_3$, and $z_5$ are linearly independent over the rational numbers $\Q$.

By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, z_5, e^{z_1}, e^{z_3}, e^{z_5}}\right)$ has transcendence degree at least $3$ over the rational numbers $\Q$.

That is, the extension field $\Q \left({\ln \ln \pi, \ln \pi, i \pi, \ln \pi, \pi, e^{i \pi}}\right)$ has transcendence degree at least $3$ over $\Q$.

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

Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$.

Therefore, $\ln \ln \pi$, $\ln \pi$, and $i \pi$ must be algebraically independent over $\Q$.

That is, $z_1$, $z_3$, and $z_5$ must be algebraically independent over rational numbers $\Q$.

It follows that $z_1$, $z_3$, $z_5$, and $1$ are linearly independent over the rational numbers $\Q$.

By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, z_5, 1, e^{z_1}, e^{z_3}, e^{z_5}, e}\right)$ has transcendence degree at least $4$ over the rational numbers $\Q$.

That is, the extension field $\Q \left({\ln \ln \pi, \ln \pi, i \pi, 1, \ln \pi, \pi, e^{i \pi}, e}\right)$ has transcendence degree at least $4$ over $\Q$.

However, $1$ is algebraic.

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

Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$.

Therefore, $\ln \ln \pi$, $\ln \pi$, $i \pi$, and $e$ must be algebraically independent over $\Q$.

That is, $z_1$, $z_3$, $z_5$, and $e$ must be algebraically independent over rational numbers $\Q$.

It follows that $z_1$, $z_3$, $e z_3$, and $z_5$ must be algebraically independent over rational numbers $\Q$.

That is, $z_1$, $z_3$, $z_4$, and $z_5$ must be algebraically independent over $\Q$.

Therefore, $z_1$, $1$, $z_3$, $z_4$, and $z_5$ must be linearly independent over $\Q$.

Hence, $z_1$, $1 + z_1$, $z_3$, $z_4$, and $z_5$ must be linearly independent over $\Q$.

That is, if Schanuel's Conjecture holds, then $z_1$, $z_2$, $z_3$, $z_4$, and $z_5$ are linearly independent over the rational numbers $\Q$.