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:
- $\paren {\ln \pi}^a \pi^b = 1$
By Schanuel's Conjecture Implies Algebraic Independence 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 $\map \Q {z_1, z_3, z_5, e^{z_1}, e^{z_3}, e^{z_5} }$ has transcendence degree at least $3$ over the rational numbers $\Q$.
That is, the extension field $\map \Q {\ln \ln \pi, \ln \pi, i \pi, \ln \pi, \pi, e^{i \pi} }$ 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 $\map \Q {z_1, z_3, z_5, 1, e^{z_1}, e^{z_3}, e^{z_5}, e}$ has transcendence degree at least $4$ over the rational numbers $\Q$.
That is, the extension field $\map \Q {\ln \ln \pi, \ln \pi, i \pi, 1, \ln \pi, \pi, e^{i \pi}, e}$ 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 $\Q$.
$\Box$