Show, for each $n$, there are only a finite number of integral $(a_i)_{i=1}^n$ such that $2\le a_i \le a_{i+1}$ and $\dfrac{\prod_{i=1}^n a_i-1}{\prod_{i=1}^n (a_i-1)} $ is an integer.
My question is inspired by this one:
Prove the fractions aren't integers
I can show that $a_1 \le \dfrac1{2^{1/n}-1} $.
I think there is a proof by induction of the general case, but I don't have one now.
A comment: This reminds me of the Egyptian fraction problem.
Here's what I have now.
$\begin{array}\\ r &=\dfrac{\prod_{i=1}^n (a_i-1)+\prod_{i=1}^n a_i-\prod_{i=1}^n (a_i-1)-1}{\prod_{i=1}^n (a_i-1)}\\ &=1+\dfrac{\prod_{i=1}^n a_i-\prod_{i=1}^n (a_i-1)-1}{\prod_{i=1}^n (a_i-1)}\\ &=1+\dfrac{\prod_{i=1}^n (b_i+1)-\prod_{i=1}^n b_i-1}{\prod_{i=1}^n b_i} \quad\text{where }b_i = a_i-1\\ \end{array} $
Letting $B = \prod_{i=1}^n b_i $, the fraction is, dividing the numerator and denominator by $B$, $r-1 =s =\prod_{i=1}^n (1+1/b_i)-1-1/B $.
We have $s \gt 1+\sum_{i=1}^n 1/b_i-1-1/B =\sum_{i=1}^n 1/b_i-1/B \gt 0 $ since $b_i \le B$ and $n \ge 2$.
Therefore, if $s < 1$, $r$ is not an integer.
If $c_i = 1/b_i$, $s = \prod_{i=1}^n (1+c_i)-1-\prod_{i=1}^n c_i $, so $\dfrac{ds}{dc_j} =\prod_{i=1, i\ne j}^n (1+c_i)-\prod_{i=1,i\ne j}^n c_i $, so that $s$ is an increasing function of each $c_i$ and so is a decreasing function of each $b_i$.
In particular, $s \le (1+1/b_1)^n-1-1/b_1^n $, so that if $(1+1/b_1)^n-1 \le 1$, $s$ is not an integer. This is $1+1/b_1 \le 2^{1/n} $ or $b_1 \ge \dfrac1{2^{1/n}-1} $.