992*. Find all natural $n$ for which the fraction $\frac{n!}{n+1}$ is reducible.
$\triangle$ Consider two cases.

1) Let the number $n-$ be odd, greater than 1.

Then the fraction is reducible by 2.

2) Let $n$ be even.

If in this case the number $n+1-$ is prime: $n+1=p$, where $p-$ is a prime, then the fraction is irreducible.

If, however, $n+1-$ is composite: $n+1=a \cdot b$, where $a$ and $b$ are natural numbers greater than 1, then the fraction is reducible, for example, by $a$.

Answer: all odd $n$, greater than 1, and all even $n \neq p-1$, where $p$ is a prime.