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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.

Adam Ledger
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    The short proof here is that $n-1\mid n^k-1$... – abiessu Oct 27 '18 at 16:15
  • sure but I really need something that enables the reader understanding in terms of the factorial's role here as simply remarking the divisibility relation you have doesn't tell us why it needs to be the factorial in order for the equality to hold true – Adam Ledger Oct 27 '18 at 16:21
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    The factorial is actually irrelevant in the equation. There could be any integer expression in place of $n!$ and the equation would not need any further proof... – abiessu Oct 27 '18 at 18:52
  • sure I can see that now in this example, I work on a lot expressions that differ only slightly and I seemed to have lost the point of what I was after in this simplification – Adam Ledger Oct 27 '18 at 19:56

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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.

John Hughes
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