Lindemann–Weierstrass theorem

For distinct algebraic numbers ${\alpha }_1,\dots ,{\alpha }_s$, the numbers $e^{{\alpha }_1},\dots ,e^{{\alpha }_s}$ are linearly independent over $\overline{\mathbb{Q}}$. Equivalently, exponentials of numbers linearly independent over $\mathbb{Q}$ are algebraically independent over $\overline{\mathbb{Q}}$. The special case of a nonzero algebraic exponent gives a transcendental exponential. In particular $e^{i\pi }=-1$ forces $\pi$ to be transcendental.

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 $k+\sum b_j{e}^{{\beta }_j}=0$, 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.