The value $e^{\gamma/(i\pi)}$ is never a root of unity, for algebraic $\gamma \neq 0$

For every non-zero algebraic $\gamma$, the value $e^{\gamma/(i\pi)}$ is not a root of unity.

Proof sketch. If $e^{\lambda}$ has finite order with $\lambda = \gamma/(i\pi)$, then $e^{n\lambda} = 1$ for some $n \geq 1$, so $n\lambda = 2\pi i m$ for some integer $m$. Substituting $\lambda = \gamma/(i\pi)$ and clearing gives $\gamma = -2\pi^{2}m/n$. With $\gamma$ algebraic and non-zero this makes $\pi^{2}$ algebraic, contradicting DiazModulus.pi_sq_transcendental, which is Proved on this mission and is discharged here rather than carried.

What it is for. DiazModulus.recip_pi_not_log and its two children ask for $e^{\gamma/(i\pi)}$ to be transcendental. This node settles the cheapest way that could fail: the value cannot be a root of unity. Anyone attacking DiazModulus.recip_pi_not_log_real_gamma needs this first, and on that half it says something concrete — $\gamma$ real makes $\lambda$ purely imaginary, so $|e^{\lambda}| = 1$, and a hypothetical algebraic value would have to be an algebraic number on the unit circle that is not a root of unity. Such numbers exist, for instance $(3+4i)/5$, so this does not close the leaf; it identifies exactly which shape a counterexample would have to take.

Not claimed. No transcendence. The gap between "not a root of unity" and "transcendental" is the whole open problem, and closing it needs transcendence input this mission does not have. This node closes nothing.

Formalization note. "Root of unity" is IsOfFinOrder applied to the value, which is the form the argument consumes; no reality or genericity hypothesis on $\gamma$ is needed, so the statement covers both axis halves at once.