given a $m$ and $n$ both natural numbers such that $1 \leq n \leq m - 1$, how do we relate the factorial $(m - n)!$ in terms of $(n - 1)!$. In other words write $(m-n)!$ in terms of $(n - 1)!$. Im having issues in understanding which terms are leftover.
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It usually helps to work an example. Suppose $m = 8$ and $n = 5$. Then $m - n = 3$. So you want to express $3!$ in terms of $4!$. If instead $n = 2$, then $m - n = 6$ and you want to express $6!$ in terms of $1!$. So either expression could be larger, so it is unclear what you are looking for. Neither is a factor of the other. For context, let's imagine you are examining ${{m - 1} \choose {n - 1}} = \frac{(m-1)!}{(n-1)!\cdot (m-n)!}$. Now it becomes clearer that the two are only related in that the sum of the two numbers in the denominator being factorialized is the number in the numerator. It is very striking to me that the expression always turns out to be an integer, not a fraction. But I suspect the relation you are looking for does not exist.
RobertTheTutor
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The reason i was looking for a relation is because i was trying to prove that $(m-n)!(n-1)! \equiv (-1)^n \mod m$ where $m$ is prime and $n$ is natural such that $1 \leq n \leq m - 1$. With wilson's theorem i know that $(m-1)! \equiv -1 \mod m$. If i could write $(m-n)!$ in terms of that then i can use wilsons theorem – UnKnoWnZ Mar 12 '21 at 05:12
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@UnKnoWnZ See the Wilson reflection formula for the modular case. – Bill Dubuque Mar 12 '21 at 08:45