How to show that $n$ is a prime? 
Suppose that $n>1$ satisfies $(n-1)! \equiv -1 \pmod n$. Show that $n$ is a prime. (Hint: Antithesis)

My own trying: $n=3$: $(3-1)!+1= 3 \cdot 1$ => $3|2!+1$. $n=5$: $(5-1)!+1=25 = 5 \cdot 5$ => $5|4!+1$. $n=7$: $(7-1)!+1=3 = 720+1 =  7 \cdot 103$ => $7|6!+1$. So at least it holds for $n=3, 5$ and $7$. But how to prove it?
 A: We will show that if  $(n-1)! \equiv -1 \pmod{n}$, and $n>1$, then $n$ is prime.
Let $a$ be any (positive) divisor of $n$ which is not equal to $n$.  Then $a \le n-1$.  So $a$ is one of the numbers which get multiplied together to form $(n-1)!$. It follows that $a$ divides $(n-1)!$.
Suppose now $(n-1)! \equiv -1\pmod{n}$.  Then, by the definition of congruence modulo $n$,  $n$ divides $(n-1)!+1$.  But since $a$ divides $n$, the fact that $n$ divides $(n-1)!+1$ implies that $a$ divides $(n-1)!+1$.
But $a$ divides $(n-1)!$.  Since $a$ divides both $(n-1)!+1$ and $(n-1)!$, it follows that $a$ divides the difference between these two numbers.  But this difference is $1$.  So $a$ divides $1$, and therefore $a=1$.
We have shown that if $a$ is any (positive) divisor of $n$ such that $a \ne n$, then $a$ must be equal to $1$.  Since $n>1$, this forces $n$ to be prime, by the definition of prime.  (The primes are precisely the numbers $n>1$ whose only positive divisors are $n$ and $1$. We have shown that if our congruence holds, and $a$ is a positive divisor of $n$ such that $a \ne n$, then $a=1$.)
A: If $n=ab$ is composite, what can you say about $a$ and $b$? Relate that to $(n-1)!$.
Here is the full answer: If $n=ab$ is a non-trivial factorization, then $a<n$ and $b<n$, and so $a\le n-1$ and $b\le n-1$. But then both $a$ and $b$ appear in $(n-1)!$ and so $(n-1)!\equiv 0 \pmod{n}$.
