Problem source

Find the set of positive integers $n$ for which $n$ does not divide $(n-1)!.$ Justify your answer. 
[Note that small values of $n$ may require special consideration.]

Solution source

Claim: $n \not \mid (n-1)!$ if and only if $n$ is prime or $4$

Proof: $(\Leftarrow)$ 
\begin{enumerate}
\item $4 \not \mid 3! = 6$. 
\item If $p$ is prime, then $p \not \mid k$ for $k < n$, therefore $p \not \mid (n-1)!$
\end{enumerate}

$(\Rightarrow)$ If $n = 1$ then $1 \mid 0! = 1$ so $1$ is not in our set. The numbers less than $6$ are all accounted for (either primes, $4$ or $1$), so let $n$ be a composite number larger than $6$, ie $n = ab$. Suppose first $a \neq b$ then $(n-1) = 1 \cdots a \cdots b \cdots (n-1)$ so $n \mid (n-1)!$. Suppose instead that $a = b$, then $n = a^2$. Since we know $a \geq 3$ we must have $1 \cdots a \cdots (2a) \cdots (a^2-1)$ so $a^2 \mid (n-1)!$ and we're done.