Prove that the sequence $ \frac{ (2^n-1)(2^{n-1}-1)...(2^{n-k+1}-1) } { (2^1-1)(2^2-1)...(2^k-1) }$ is an integer For any given integer $n\ge 1$ and $k \in\{1, 2, \dots, n\},$
$$ F_n(k) = \frac{ (2^n-1)(2^{n-1}-1)...(2^{n-k+1}-1)   } { (2^1-1)(2^2-1)...(2^k-1) } $$
For example, if $n = 1,$ then $k \in \{1\}$ and $$F_1(1) = \frac{2^1-1}{2^1-1} $$
For $n=3,$ then $k\in\{1, 2, 3\}$ and
$$ F_3(1) = \frac{2^3-1}{2^1-1}  = 7 $$
$$ F_3(2) = \frac{(2^3-1)(2^2-1)}{(2^1-1)(2^2-1)} =7 $$
$$ F_3(3) = \frac{(2^3-1)(2^2-1)(2^1-1)}{(2^1-1)(2^2-1)(2^3-1)} =1 $$

Cobsider $(n)_q:=\frac{q^n-1}{q-1} $, the so called q-analogue of $n$. Observe that this is a polynomial of degree n-1 in the variable q.
Define now the q-factorial and the q-binomial as follow:
$$(n)_q!:= \prod_{i=1}^n (i)_q  \qquad \binom{n}{k}_q:=\frac{(n)_q!}{(k)_q! \,(n-k)_q!}$$
It is easy to prove that both are polynomials, for example using the identity
$$
\binom{n}{ k}_q = q^k \binom{n-1}{k}_q + \binom{n-1}{k-1}_q
$$
What you need follows from the evaluation in 2 on the q-binomial.
