10 years later the author of that proof wrote a little book, "Irrational Numbers", that contains a slightly more powerful form of that proof. It proves that pi^2 is irrational.
It's basically the same proof except with the following changes.
1. Instead of f(x) = x^n (a - bx)^n / n!, where pi = a/b, it defines f(x) = x^n (1-x)^n / n!. It notes that for 0 < x < 1 we have 0 < f(x) < 1/n!.
2. It uses a lemma that was proven a few pages earlier to show that the j'th derivative of f at 0, f^(j)(0), is an integer for all j. Here's the lemma and proof [1].
3. It notes that f(x) = f(1-x), and so f^(j)(1) is also an integer for every j.
4. As in the pi irrationality proof, it defines F(x) as an alternating series of derivatives of f(x), but the F(x) is a little different:
5. It takes the derivative of F'(x) sin(pi x) - pi F(x) cos(pi x), giving pi^2 a^n f(x) sin(pi x).
6. Integrate that divided by pi from 0 to 1 to give pi a^n integral(f(x) sin(pi x), 0, 1) = F(1) + F(0).
7. Note that F(1) + F(0) is an integer.
8. From 0 < f(x) < 1/n! (see item #1 above), we have that 0 < pi a^n integral(f(x) sin(pi x), 0, 1) < pi a^n/n! < 1 for sufficiently large n, contradicting item #7.
[1] Lemma: if h(x) = x^n g(x) / n! where g(x) is a polynomial with integer coefficients, then h^(j)(0) is an integer for j = 0, 1, 2, ..., and with the possible exception of the case j = n that integer is divisible by n+1. No exception is needed in the case j = n if g(0) = 0.
Proof: h^(j)(0) = Cj j! / n! where Cj is the coefficient of x^j in x^n g(x) so Cj is an integer by hypothesis. (That's all that is needed in the pi^2 irrationality proof, so you can stop here if that's all you are interested in).
For j < n we have Cj = 0 so h^(j)(0) = 0 which is divisible by n + 1.
For j > n we have a factor of n + 1 in j! / n!, so h^(j)(0) will also have a factor of n + 1.
For j = n, we have h^(j)(0) = Cn. If g(0) = 0, then Cn = 0, which is divisible by n + 1.
If one wants to get it from there I'd recommend either the 6.2 MB PDF of the 6th printing, or the 8.0 MB PDF that doesn't state the printing in the description, but is the only 8.0 MB PDF.
Both of those are searchable PDFs, and pretty good scan quality. The 8.0 MB one has the best scan quality, although the pages are a little yellowed which might be a bit annoying.
The 8.0 MB one is actually just a copy of the Internet Archive's scan, which is available for borrowing directly from the IA [1].
There's a 3.9 MB PDF but the scan quality is noticeably worse than the others and it is not searchable [2].
I've got a searchable PDF I made myself from my physical copy of the book using a CZUR ET24 Pro. The IA scan text is little better, but not by much, and their file is a little smaller (mine is 9.1 MB), but overall I like my scan more. I've been quite impressed with the CZUR scanner.
[2] By "not searchable" I mean it doesn't appear to have text information included to allow for fast search. Some PDF readers can still search it but I think they may do their own OCR on it to do so, and that takes a while. The ones I've said are searchable can search the whole document almost instantly.
It's basically the same proof except with the following changes.
1. Instead of f(x) = x^n (a - bx)^n / n!, where pi = a/b, it defines f(x) = x^n (1-x)^n / n!. It notes that for 0 < x < 1 we have 0 < f(x) < 1/n!.
2. It uses a lemma that was proven a few pages earlier to show that the j'th derivative of f at 0, f^(j)(0), is an integer for all j. Here's the lemma and proof [1].
3. It notes that f(x) = f(1-x), and so f^(j)(1) is also an integer for every j.
4. As in the pi irrationality proof, it defines F(x) as an alternating series of derivatives of f(x), but the F(x) is a little different:
F(x) = b^n { pi^(2n) f(x) - pi^(2n-2) f^(2)(x) + pi^(2n-4) f^(4)(x) - ... + (-1)^n f^(2n)(x)}
where pi^2 = a/b.
5. It takes the derivative of F'(x) sin(pi x) - pi F(x) cos(pi x), giving pi^2 a^n f(x) sin(pi x).
6. Integrate that divided by pi from 0 to 1 to give pi a^n integral(f(x) sin(pi x), 0, 1) = F(1) + F(0).
7. Note that F(1) + F(0) is an integer.
8. From 0 < f(x) < 1/n! (see item #1 above), we have that 0 < pi a^n integral(f(x) sin(pi x), 0, 1) < pi a^n/n! < 1 for sufficiently large n, contradicting item #7.
[1] Lemma: if h(x) = x^n g(x) / n! where g(x) is a polynomial with integer coefficients, then h^(j)(0) is an integer for j = 0, 1, 2, ..., and with the possible exception of the case j = n that integer is divisible by n+1. No exception is needed in the case j = n if g(0) = 0.
Proof: h^(j)(0) = Cj j! / n! where Cj is the coefficient of x^j in x^n g(x) so Cj is an integer by hypothesis. (That's all that is needed in the pi^2 irrationality proof, so you can stop here if that's all you are interested in).
For j < n we have Cj = 0 so h^(j)(0) = 0 which is divisible by n + 1.
For j > n we have a factor of n + 1 in j! / n!, so h^(j)(0) will also have a factor of n + 1.
For j = n, we have h^(j)(0) = Cn. If g(0) = 0, then Cn = 0, which is divisible by n + 1.