if $n>5$ is prime,prove $(n-1)|(n-2)!$ Could you give a hint how to prove this?
if $n>5$ is prime, prove $(n-1)|(n-2)!$
 A: This is not the same proof as before, I have changed the proof to the following:
If $n > 5$ is prime then $n-1$ is composite as there are no consecutive primes in this case. Hence, $n-1 = ab$ such that $1 < a < b < n-2 < n-1$ or $n-1 = p^{2}$ where $p$ is prime. For the first case this means that $a,b \in \{2,...n-2\}$ so that $ab|(n-2)!$ and $(n-1)|(n-2)!$. If $n-1 = p^{2}$, in order to get both $p$'s in the expansion of $(n-2)!$, we need to get $(n-2) \geq 2p = 2\sqrt{n-1}$ since if we consider the set $\{1,2,...,p,...2p\}$, this is the smallest set that when one takes the product, we have that $p^{2}$ divides the result. The previous inequality is exactly true for all $n>7$, but more specifically when $n$ prime.
Therefore in either case, $(n-1)|(n-2)!$.
A: If $n$ is prime, then $n-1$ is composite unless $n=2 \text{ or } 3$. So the following line of argument begins by assuming $n>3$.
If $n>3$, then $n-1$ is composite, and it can be factored into two smaller integers, $(n-1)=ab$, each of which is smaller than $n-2$. This is true because an integer multiple of any positive integer $>1$ cannot equal the next larger integer (i.e. $k(n-2)\ne n-1$), so neither of $a,b$ can be as large as $(n-2)$.
If $n-1$ can be factored such that $a\ne b$, since $a,b<n-2$, they will each appear as separate terms in the product $(n-2)!$, and we are done. $a\mid (n-2)!$ and $b\mid (n-2)!$, so $ab\mid (n-2)!$, meaning $(n-1)\mid (n-2)!$. For odd primes, $n-1$ is even and can be always factored as $2$ times another number different from $2$, except when $n=5$. So far, the argument has been made assuming $n>3$, but now we must consider $5$ as a special case.
For $n=5$, where $n-1=4=2\times 2$ and $n-2=3$, the terms $1,2,3$ of $(n-2)!$ contain a factor of $2$ only once. In fact, $4\not \mid 3!=6$, so the proof fails for the case $n=5$, and we must require $n>5$ for it to be valid.
I hope that this not only shows that the proposition is true, but also explains why it must be conditioned on $n>5$
