If e^{π²} is algebraic, then e and π are algebraically independent

If $e^{\pi^{2}}$ is algebraic, then $e$ and $\pi$ are algebraically independent over $\mathbb{Q}$.

The transcendence of $e^{\pi^{2}}$ itself is not known. What is known unconditionally is that $e^{\pi^{2}}$ or $e^{i\pi^{3}}$ is transcendental (DiazModulus.exp_pi_sq_or_exp_i_pi_cube_transcendental).

Proof. Apply DiazModulus.two_algebraically_independent_of_exp_column at $x = (i\pi, 1)$ and $y = (1, i\pi)$. The column $y_2 = i\pi$ carries $e^{-\pi^{2}}$ and $e^{i\pi} = -1$, both algebraic. All eight numbers are then algebraic over $\mathbb{Q}(\pi, e)$, so that field has transcendence degree two: $\pi$ and $e$ are algebraically independent.

Novelty. None: this is Corollaire 1 of Waldschmidt (1973) at $\alpha = -1$ (with $\log\alpha = i\pi$), and the paper states it in the remark that follows the corollary (p. 192). The substitution is the one in the corollary's proof (p. 193). The contribution of this node is the formal proof.