TITLE: Transcendence of PI
QUESTION [7 upvotes]: Can anyone suggest me an ingenious proof of the transcendence of $\pi$. I have seen Lindemann's proof but it appears intricate.

REPLY [7 votes]: An unfortunate thing is that proving the transcendence of $\pi$ has the same complexity as the Lindemann--Weierstrass theorem. (Proving the transcendence of $e$ is much easier.) I am quite surprised to see that the book 
[A.I. Galochkin, Yu.N. Nesterenko, and A.B. Shidlovskiĭ,
Введение в теорию чисел [Introduction to number theory],
2nd edition, Moscow State Univ., 1995. 160 pp. ISBN: 5-211-03075-3. MR1367734 (96i:11001)] is not translated into English (I remember that there was an attempt to negotiate the translation about 10 years ago). I put here an extract from Yu. Nesterenko's lectures (in Russian!) on the Lindemann--Weierstrass theorem which are "cleaned", so I believe this is the simplest version of proof.
There are more exotic ways to state the transcendence of $\pi$, of course, these proofs are more involved. One example in this direction is [V.N. Sorokin, On the measure of transcendency of the number $\pi^2$, Sb. Math. 187  (1996),  no. 12, 1819--1852] where the author shows the transcendence of $\pi^2$ (rather than $\pi$ itself) by constructing linear forms involving the numbers
$$
\sum_{n_1\ge n_2\ge\dots\ge n_r\ge1}\frac1{n_1^2n_2^2\cdots n_r^2}
$$
which are (now) known to be simple rational multiple of $\pi^{2r}$, $r=1,2,\dots$.