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Problem: If $n>4$ and is composite, then $n$ divides $(n-1)!$.

My Attempt: I have tried to solve this problem using the approach described by @Michael Hardy but have not succeeded as of now. I wish to know how one can solve this problem using his approach. For reference, Michael's answer is given below:

Suppose $p$ is prime and $p\mid n$. Let $k$ be the multiplicity of $p$ as a factor of $n$, so that $p^k\mid n$. $$ \underbrace{(n-1)! = 1\cdot2\cdot3\cdot4\cdot5\cdots\cdots p\cdots\cdots p \cdots\cdots p\cdots\cdots(n-1)}_{\text{Does $p$ show up at least $k$ times here?}} $$If $p$ appears at least $k$ times in the usual factorization of $(n-1)!$ displayed above, then $p^k\mid(n-1)!$. If this works for every prime factor of $n$, then they're all there, so $n\mid(n-1)!\,{}$.

I guess, the question really boils down to proving the fact that the prime number $p$ shows up at least $k$ times in $(n-1)!$.

Student
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  • You should try writing $n$ as its factorization: then each factor appears as a term of the product $(n-1)!$. – Riccardo Orlando Nov 04 '16 at 17:59
  • @RiccardoOrlando multiplicand. – Jacob Wakem Nov 04 '16 at 18:00
  • @WimC I have read all other posts before asking this question. My question is different because it asks for a proof using an approach which I have mentioned. Please use that approach to come up with a rigorous proof. – Student Nov 04 '16 at 18:04
  • @Shrey Did you intend to restrict your question to that particular method of proof? Please give a link to the proof you cite. – Bill Dubuque Nov 04 '16 at 18:48
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    http://math.stackexchange.com/questions/718609/prove-that-if-n-is-composite-then-n-1-equiv-0-pmod-n?noredirect=1&lq=1 – Student Nov 04 '16 at 19:03
  • If you don't get a satisfactory answer in the comments there then please edit your question to make it clear that it is specific to this method and then ping me and I will reopen it. – Bill Dubuque Nov 04 '16 at 19:18

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Well consider some prime $p$, and the largest integer $k$ such that $p^k\mid n$. We know that all numbers that have $p$ as a factor are of the form $p$, $2p$, $3p$, $\dots$. Or more generally of the form $mp$ for $m\in\mathbb{N}$. Thus if there are $k$ factors of $p$ in $n$, we know that the first $k$ numbers with $p$ as a factor are $p,\dots,kp$. However we know that $kp< p^k\le n$ (in our scenario of $n>4$ and $n$ composite atleast, I'll leave this proof up to you) thus all the numbers $p,\dots, kp$ are less than $n$ and will occur as factors in the product $(n-1)!=2\cdot 3\cdot\dots\cdot (n-1)$.

Will Fisher
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