= Symmetric Hermite integral obstruction
{title2=$k+\sum_jb_je^{\beta_j}\ne0$}
Suppose nonzero <algebraic numbers> $\beta_j$ form a Galois-stable set, with Galois-invariant <integer> weights $b_j$, and $k\ne0$ is an <integer>. Choose $P\in\mathbb Z[X]$ vanishing at all $\beta_j$ with $P(0)\ne0$, and a positive <integer> $c$ making all $c\beta_j$ <algebraic integers>. For a large prime $p$, put $D=(1+\deg P)p-1$ and $f_p(X)=c^DX^{p-1}P(X)^p/(p-1)!$, and let $F_p$ be the sum of its derivatives. The weighted sum $kF_p(0)+\sum b_jF_p(\beta_j)$ is an <integer>, nonzero modulo $p$ when $p\nmid kcP(0)$. If the displayed exponential expression vanished, integrating $(e^{-z}F_p(z))'=-e^{-z}f_p(z)$ along the finitely many segments to $\beta_j$ would bound that nonzero <integer> by $C^p/(p-1)!\to0$. This contradiction proves the obstruction and supplies direct proofs for $e$ and $\pi$.
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