And the formula of all rational products seems to suggest that taking some n as n approaches infinity, the formula will have an always increasing amount of uncancelled primes(so provably non telescoping) in the total resulting product. Is that correct?
Taken from formula 48 in http://mathworld.wolfram.com/InfiniteProduct.html.
$$\prod_{k=1}^\infty (1+\frac{2}{k})^{-1^{k+1}*k} = \frac{\pi }{2e} $$
It has products of the form
$$ \left(\frac{3}{1}\right)^1 \left(\frac{1}{2}\right)^2 \left(\frac{5}{3}\right)^3 \left(\frac{2}{3}\right)^4 \left(\frac{7}{5}\right)^5 \left(\frac{3}{4}\right)^6 \left(\frac{9}{7}\right)^7 \left(\frac{4}{5}\right)^8 \left(\frac{11}{9}\right)^9 \left(\frac{5}{6}\right)^{10}\left(\frac{13}{11}\right)^{11}$$
Due to the $-1^{k+1}$ you get $\left(\frac{k+2}{k}\right)^{k}$ for odd k, and its obvious this will take on every odd number (and obviously every odd prime) to the $k^{th}$ power.
You can define a function F(P) that is equal to the uncancelled terms formed by this product(and thus equal to $\frac{\pi }{2e} $ iff it is a rational number), and show some properties of it.
In the expansion of the product, because the power of each term is increasing; so when a given k+2 is prime, the next term cannot fully cancel it, and because its prime and each term is only one higher than the previous in terms of the unreduced fraction, you have to wait for 2k+4 times it. Since tt is known that their is always a prime between x and 2*x for any x>1; then F(P) as P -> infinity tends towards having an arbitrarily large set of non cancelable factors in its fraction expression, which contradicts it being a rational number.
This does not prove that $\pi/e$ is irrational. You can have a rational number $r$ as a limit of a sequence $s_n$ of rationals where each $s_n$ has a denominator divisible by a prime $p$, but $r$ does not. For example, $1 = \lim_{n \to \infty} \dfrac{2^n-1}{2^n}$.