TITLE: Proof that pi is transcendental that doesn't use the infinitude of primes
QUESTION [58 upvotes]: I just taught the classical impossible constructions for the first time, and in finding my class a reference for the transcendence of pi, I found a dearth of distinct proofs.  In particular, those that I read all require the existence of infinitely many primes, which strikes me as extraneous.  Is there a known proof that requires only knowledge that I would "expect", namely, integral calculus to get your hands on the actual constant and algebraic properties of polynomials in connection with the assumption that the constant is algebraic?

REPLY [18 votes]: Barry, this post is coming a "little bit" after your class ended, but I want to direct you to a paper on the Lindemann-Weierstrass theorem by Beukers, Bezivin, and Robba in the Amer. Math. Monthly in 1990: https://www.jstor.org/stable/pdf/2324683.pdf.  They give a proof of LW theorem that does not explicitly involve a choice of an auxiliary prime number, and it would be a good exercise to work out what the proof is saying in the special case of the number $\pi$ to see more directly how it proves $\pi$ is transcendental, or more generally how it shows $e^\alpha$ is transcendental when $\alpha$ is a nonzero algebraic number (the Lindemann part of LW) without dealing with the more general setting of a finite set of algebraic $\alpha_i$ (the Weierstrass part of LW). As indicated at the start of the paper by Beukers et al., the proof was inspired by ideas from $p$-adic analysis, but their paper is a simplified version that does not have any explicit dependence on $p$-adic notions.
I don't agree with the post claiming infinitude of primes "might be necessary" to prove transcendence of $\pi$. Just because you are working within an axiomatic framework that would allow you to prove infinitude of primes doesn't mean you ought to be required to do that for what you want, so that whole matter seems irrelevant to answering your question. If I am proving the Eisenstein irreducibility criterion in $\mathbf Z[x]$ you can't seriously tell me that I must go on a detour in the middle of the proof and prove $\mathbf Z[i]$ is a Euclidean domain just because it is logically possible with the axioms I am allowing. In any case, I think the paper I linked to should settle your question in an affirmative way that you were seeking.
UPDATE: The impression that you need infinitely many primes to prove transcendence results is incorrect, not because there are proofs that avoid it like the comparatively recent proof I mention above or older proofs I did not mention, but because even the proofs that mention it don't really need it. For example, look at the two proofs of transcendence of $e$ and $\pi$ here (pp. 3-5) and here. (In case those links change in the future, the proofs are essentially the same as those on pp. 4-6 of Baker's book "Transcendental Number Theory".) The way the proofs work is that an expression $J_p$ is introduced that depends on a prime $p$ (and some fixed data like a hypothetical algebraic relation for the number to be proved transcendental) and $J_p$ is shown to satisfy two properties: (i) a growth estimate $|J_p| \leq c^p$, where $c$ is independent of $p$ and (ii) $J_p$ is a sum of two integers, one being a multiple of $p!$ and the other being a multiple of $(p-1)!$ that for all large prime $p$ is not a multiple of $p!$, so $J_p$ is an integer multiple of $(p-1)$! that is not zero. Therefore $|J_p| \geq (p-1)!$, which contradicts the growth estimate in (i) since we can't have $(m-1)! \leq c^m$ for large $m$. How is the primality of $p$ really used? Its only purpose is to be an arbitrarily large number that is relatively prime to a couple of specific nonzero integers. But there is no need to rely on prime numbers to achieve that: if you want a huge integer $m$ relatively prime to two nonzero integers $a$ and $b$, just take $m = |ab|M + 1$ for a huge integer $M$. So you can work with numbers in such an arithmetic progression in place of a large prime number $p$ in the proof. (Personally I'd also pass to the integer $J_m/(m-1)!$ so the bounds $0 < |J_m/(m-1)!| \leq c^m/(m-1)!$ lead to the "no integers between $0$ and $1$" contradiction once $c^m/(m-1)!< 1$, but that's a matter of taste.) 
Of course the idea of creating numbers relatively prime to particular numbers by taking a multiple of their product and adding $1$ is the main idea in Euclid's proof of the infinitude of the primes, so we aren't bypassing a method that can lead to infinitely many primes but we are bypassing any need for prime numbers in the proof that uses them.
The idea of using a large prime in proofs of the transcendence of $e$ and $\pi$ is due to Hurwitz, who added this detail to a proof by Hilbert that relies on divisibility properties but not on primes.  Mahler discusses this in his Springer Lecture Notes in Math "Lectures on Transcendental Numbers" (1976). The volume is available online here, and while most chapters are behind a paywall the last part called Back Matter is not, and that's the part relevant to us. Mahler presents Hilbert's proof starting on p. 237. On p. 243 he mentions Hurwitz's idea of using a prime instead of a number in a simply constructed arithmetic progression and explains why he thinks this "seemingly simpler" approach of Hurwitz is actually not a good idea.