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?

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 $(n-1)!\equiv 0 \pmod{n}$.
