Prove that $n !$ is a divisor of $ \prod_{k=0}^{n-1}\left(2^{n}-2^{k}\right) $

Question

Prove that for any natural number $n, n !$ is a divisor of $ \prod_{k=0}^{n-1}\left(2^{n}-2^{k}\right) $

i have already seen it here $\prod_{i=0}^{n-1}(2^n-2^i)$ can be divided by $n!$ but my doubt is different.

Solution -

for $p=2$ we can easily prove that $v_{2}\left(\prod_{k=0}^{n-1}\left(2^{n}-2^{k}\right)\right)=\sum_{k=0}^{n-1} v_{2}\left(2^{n}-2^{k}\right) \geq n-1$ > $n-S_{2}(n)=e_{2}(n)$

let now $p>2$ ,and author proves that

$ v_{p}\left(\prod_{k=0}^{n-1}\left(2^{n}-2^{k}\right)\right) \geq\left\lfloor\frac{n}{p-1}\right\rfloor $ ( using FLT)

and

$ e_{p}(n)=\frac{n-s_{p}(n)}{p-1} \leq \frac{n-1}{p-1}<\frac{n}{p-1} $ and since $e_{p}(n)$ is an integer, we must have $ e_{p}(n) \leq\left\lfloor\frac{n}{p-1}\right\rfloor $

now i did not get this last step. why $ e_{p}(n) \leq\left\lfloor\frac{n}{p-1}\right\rfloor $ we know that $x>[x]$ so how we get reversed signs ???

thankyou

Answer 1

If you have an ineteger $n<5.5$ .Then it implies $n\le \lfloor5.5\rfloor=5$

This is the same thing.Since $e_p(n) < \frac{n}{p-1}$ and $e_p(n)$ is an integer .We have that, $e_p(n) \le \lfloor \frac{n}{p-1}\rfloor$

Answer 2

You know that $e_{p}(n)$ is an integer smaller than $\frac{n}{p-1}$.

At the same time $\left\lfloor\frac{n}{p-1}\right\rfloor$ is the largest integer smaller than $\frac{n}{p-1}$.