Worked solution: Lindemann's transcendence proof of $\pi$ settles squaring the circle (1882)

Lindemann's 1882 paper 'Über die Zahl $\pi$' proves: for every nonzero algebraic number $\alpha$, the value $e^\alpha$ is transcendental. This generalizes Hermite's 1873 result (the case $\alpha=1$, giving $e=e^1$ transcendental) to arbitrary algebraic exponents, including complex ones.

The proof adapts Hermite's auxiliary-integral construction, but where Hermite worked with a single integer $\alpha=1$, Lindemann must work simultaneously with $\alpha$ and all of its algebraic conjugates (the other roots of $\alpha$'s minimal polynomial), multiplying auxiliary integrals over the whole set so that the symmetric-function combinations appearing in the final estimate are guaranteed to be rational — a substantially harder bookkeeping problem that Lindemann resolved using elementary symmetric polynomials and the same two-sided estimate (forced nonzero integer vs. forced arbitrarily small) as Hermite's argument.

Applying this theorem to $\alpha=i\pi$ (assumed nonzero and algebraic in Step 3) gives immediately that $e^{i\pi}$ must be transcendental. But Step 3 also showed $e^{i\pi}=-1$, which is algebraic — a direct contradiction. The only way out is that the assumption was false: $\pi$ is not algebraic, i.e. $\pi$ is transcendental (Lindemann 1882).