For any prime $p$, $p^2 \nmid p!$ as prime $p$ occurs only once in the prime factorization of $p!$. Additionally, $4^2 = 16$ does not divide $4! = 24$. We will show that $n^2 \mid n!$ for all other positive integers $n$. Let $p$ be a prime factor of $n$, and $k$ the exponent of $p$ in the prime factorization of $n$. If there exists a different prime factor $q$ of $n$, then both $p^k$ and $p^kq$ are in the set of integers from 1 to $n$ and hence $p^{2k} \mid n!$. This holds for all prime factors $p$ of $n$.

Thus $n^2$ divides $n!$ for all $n$ with at least two distinct prime factors. Now consider the remaining integers $n = p^k$ for prime $p$. For $k \ge 3$ all three integers $p$, $p^{k-1}$, and $p^k$ are distinct factors in $n!$, implying that $p^{2k} = n^2$ divides $n!$. For $k = 2$ and $p > 2$ the integers $p$, $2p$, and $p^2$ are distinct factors in $n!$, implying that $p^4 = n^2$ divides $n!$.

We find all positive integers $n$ such that $n \mid (n-1)!$; obviously this condition is equivalent to that of the problem. For prime $n$, no integer less than $n$ can have a prime factor $n$ and thus $n$ cannot divide $(n-1)!$. For $n = 4 = 2^2$, 4 does not divide $(n-1)! = 6$. On the other hand, for $n = p^2$ where $p > 2$ is prime, the integers $p$ and $2p$ are in the set of integers from 1 to $n-1$, implying $p^2 = n \mid (n-1)!$. If $n$ is a cube or a higher power of some prime, or has at least two prime factors, then it obviously has a divisor $d$ such that $1 < d < n$ and $d^2 \neq n$ (e.g. the smallest prime factor of $n$). Hence $d$ and $\frac{n}{d}$ are distinct positive integers less than $n$ and their product $n$ divides $(n-1)!$.