= Hermite auxiliary-polynomial proof for pi
{c}
Under an assumed algebraicity of $i\pi$, form the product $\prod_j(1+e^{\alpha_j})$ over its conjugates; its one known vanishing factor makes the whole product zero and regroup the subset sums as $k+\sum b_je^{\beta_j}=0$, with nonzero exponents and Galois-invariant integral weights. After clearing their denominators, use $f_p(z)=(cz)^{p-1}\prod_j(cz-c\beta_j)^p/(p-1)!$. The sum of its derivatives at each $\beta_j$ is divisible by $p$ in the algebraic integers, whereas at zero its first nonzero derivative is a fixed nonzero residue for all but finitely many primes. Summing with the weights gives a nonzero rational integer. Integrating the derivative of $e^{-z}\sum f_p^{(j)}(z)$ expresses that integer as a quantity bounded by $C^p/(p-1)!$, which tends to zero. This is the contradiction.
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