The product of $\frac{b^i-a}{b^i-1}$ lies in a special ring (conjecture)

Question

Consider any three positive integers $a, b, n$. Is it true that $$\frac{(b-a)(b^2-a)\cdot\dotsb\cdot(b^n-a)}{(b-1)(b^2-1)\cdot\dotsb\cdot(b^n-1)}\in\mathbb{Z}\left[\frac{a-1}{b-1},\frac{a^2-1}{b^2-1},\dotsc ,\frac{a^n-1}{b^n-1}, 1/2 \right]?$$

Moreover, if the first question is true, then what is the minimal value of the degree $\deg_{t_{n+1}}$ of the polynomial $F(t_1, t_2,\ldots , t_{n+1})$ with $n+1$ variables and the integer coefficients, for which $$\frac{(b-a)(b^2-a)\cdot\dotsb\cdot(b^n-a)}{(b-1)(b^2-1)\cdot\dotsb\cdot(b^n-1)}=F\left(\frac{a-1}{b-1},\frac{a^2-1}{b^2-1},\dotsc,\frac{a^n-1}{b^n-1}, 1/2 \right)?$$

This conjecture is a generalization of my post here. And I will be interested in any comments on it.

Answer 1

This may be understood $p$-adically as is done in my answer to your question on math.SE. It would be nice to see the algebraic formula proving the same statement (that is, the explicit polynomial not depending on $a$ and $b$ with integer coefficients for which $$ P\left(b,\frac{a-1}{b-1},\dots,\frac{a^n-1}{b^n-1}\right)=2^{[n/2]+[n/4]+\dots}\cdot \frac{(b-a)(b^2-a)\dots (b^n-a)}{(b-1)(b^2-1)\dots(b^n-a)}. $$