At this point in my ongoing study of a Prime number relation I posted as a question two months ago requires me to find a direct proof for the following:
$\forall n \gt 2 \land k \gt 1 $
$$n!= \Biggl\lfloor { \frac {n!\, \left( {n}^{k}-1 \right) ^{n}}{ \left( n-1 \right) ^{n}}} \Biggr\rfloor \frac{\left( n-1 \right) ^{n}}{ \left( {n}^{k}-1 \right) ^{n} }$$
I already have a very long winded proof involving a Discrete Fourier Transform that people will not like, but I am hoping to find a simpler more elementary proof.
Because $n-1$ divides evenly into $n^k - 1$ (the quotient is $n^{k-1} + n^{k-2} + \ldots + 1$, because of the geometric series theorem), the number $$ u = \frac{ (n^k - 1)^n}{(n-1)^n} = \left(\frac{n^{k}-1}{n-1}\right)^n $$ is an integer, hence $n!u$ is an integer as well, and since the floor of an integer is that integer, your expression on the right becomes $$ \lfloor n! u \rfloor \cdot \frac{1}{u} = (n!u) \frac{1}{u} = n!, $$ and your claim is proved.