I try to prove that for every positive integers $m\ge n$, the following product is an integer: $$\prod_{k=0}^{n-1}\frac{2^{2^m}-2^{2^k}}{2^{2^n}-2^{2^k}}.$$
But no luck.
I try to prove that for every positive integers $m\ge n$, the following product is an integer: $$\prod_{k=0}^{n-1}\frac{2^{2^m}-2^{2^k}}{2^{2^n}-2^{2^k}}.$$
But no luck.
The fraction equals $$\prod_{k=0}^{n-1}\frac{2^{2^m-2^k}-1}{2^{2^n-2^k}-1},$$ hence it suffices to show that $$\prod_{k=0}^{n-1}\frac{x^{2^m-2^k}-1}{x^{2^n-2^k}-1}\in\mathbb{Z}[x].$$ The numerator and the denominator factor into cyclotomic polynomials. If $a(d)$ and $b(d)$ are the multiplicties of the $d$-th cyclotomic polynomial in the numerator and the denominator, then it suffices to show that $a(d)\geq b(d)$. Now $a(d)$ equals the number of $k\in\{0,\dotsc,n-1\}$ such that $d\mid 2^m-2^k$, while $b(d)$ equals the number of $k\in\{0,\dotsc,n-1\}$ such that $d\mid 2^n-2^k$.
Let us write $d=2^r e$ with $e$ odd. If $r\geq n$, then $a(d)=b(d)=0$, and we are done. Assume now that $r<n$. Then $a(d)$ is the number of $k\in\{r,\dotsc,n-1\}$ such that $e\mid 2^{m-k}-1$, while $b(d)$ is the number of $k\in\{r,\dotsc,n-1\}$ such that $e\mid 2^{n-k}-1$. Let $s$ be the multiplicative order of $2$ modulo $e$. Then $a(d)$ is the number of elements of $\{m-n+1,\dotsc,m-r\}$ divisible by $s$, while $b(d)$ is the number of elements of $\{1,\dotsc,n-r\}$ divisible by $s$. Hence $$a(d)=\left\lfloor\frac{m-r}{s}\right\rfloor-\left\lfloor\frac{m-n}{s}\right\rfloor,\qquad b(d)=\left\lfloor\frac{n-r}{s}\right\rfloor,$$ and we are left with proving $$\left\lfloor\frac{m-r}{s}\right\rfloor\geq\left\lfloor\frac{m-n}{s}\right\rfloor+\left\lfloor\frac{n-r}{s}\right\rfloor.$$ However, this is clear from the inequality $\lfloor u+v\rfloor\geq\lfloor u\rfloor+\lfloor v\rfloor$, and we are done.