2. If $(n-1)$ ! is not divisible by $n^{2}$, find all possible values of $n$.

If $(n-1)$ ! cannot be divided by $n^{2}$, find all possible $n$.
2. Solution: If $n$ is a prime number, then obviously $n^{2} \times (n-1)!$. If $n$ is not a prime number.
(1) Let $n=ab$, where $a$ and $b$ are coprime integers greater than 2, then $(n-1)!$ contains factors $a, 2a, b, 2b$, so $n^{2} \mid (n-1)!$.
(2) Let $n=p^{2}$, where $p$ is a prime number greater than 5, then $p^{2}-1 > 4p$, hence $(n-1)! = (p^{2}-1)!$ contains factors $p, 2p, 3p, 4p$, thus it can be divided by $p^{4}=n^{2}$.
(3) Let $n=2p$, where $p$ is an odd prime number, then $n^{2} \times (n-1)!$.
(4) When $n=4,8,9$, in all cases, $n^{2} \times (n-1)!$,
Therefore, the required $n=8,9, p, 2p$ ($p$ is a prime number).