How prove $n|2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)$

Question

Question:

Today, when I solve other problem, I found this follow interesting result

$$n\mid\left(2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)\right),n\ge 1$$

It is clear $n=1$ is true. $n=2$

$$2\mid2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)=2$$

when $n=3$.then $$3\mid2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)=8\cdot 3=24$$

when $n=4$,then $$4\mid64\cdot 3\cdot 7$$ $$\cdots,\cdots,$$

then How prove this general?

I know this Euler theorem $$a^{n-1}-1\equiv 1\pmod n,(a,n)=1$$ How prove it?

Answer 1

Let $n=2^st$ where $t$ is odd. If $t>1$ then $$t\mid 2^{\phi(t)}-1\ ,$$ and $\phi(t)$ is one of the numbers $1,2,\ldots,n-1$, so $$t\mid (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)\ ;$$ obviously this is also true when $t=1$. Also, $s<n\le\frac{1}{2}n(n-1)$, so $$2^s\mid 2^{n(n-1)/2}\ ,$$ and hence $$n=2^st\mid 2^{n(n-1)/2}(2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)$$ as claimed.