 Symmetric Hermite integral obstruction

Suppose nonzero algebraic numbers ${\beta }_j$ form a Galois-stable set, with Galois-invariant integer weights $b_j$, and $k\ne 0$ is an integer. Choose $P\in \mathbb{Z}[X]$ vanishing at all ${\beta }_j$ with $P(0)\ne 0$, and a positive integer $c$ making all $c{\beta }_j$ algebraic integers. For a large prime $p$, put $D=(1+\mathrm{deg}P)p-1$ and $f_p(X)=c^D{X}^{p-1}P(X{)}^p/(p-1)!$, and let $F_p$ be the sum of its derivatives. The weighted sum $k{F}_p(0)+\sum b_j{F}_p({\beta }_j)$ is an integer, nonzero modulo $p$ when $p\nmid kcP(0)$. If the displayed exponential expression vanished, integrating $(e^{-z}{F}_p(z){)}^{\prime }=-e^{-z}{f}_p(z)$ along the finitely many segments to ${\beta }_j$ would bound that nonzero integer by $C^p/(p-1)!\to 0$. This contradiction proves the obstruction and supplies direct proofs for $e$ and $\pi$.