Calculating that 13 is a Wilson Prime I'm confused by the idea of a Wilson Prime. The theorem states that $$p^2=(p-1)!+1$$
This makes sense for $5$: $$5^2=(4\times3\times2)+1$$  so $5^2=25$
But it makes no sense to me for $13$: $$13^2=4790016001$$
Clearly I am as far from a mathematician as possible. If you can help in very simple terms...
 A: A Wilson prime is a prime $p$ such that $p^2$ divides $(p-1)!+1$. 
So you need to check that $169$ divides $12!+1$.
Remark: By Wilson's Theorem (called that despite the fact it was never proved by Wilson) $(p-1)!+1$ is always divisible by $p$ if $p$ is prime. The prime $p$ is called a Wilson prime if $(p-1)!+1$ has at least one "extra" factor of $p$. 
A: To prove that $13$ is a Wilson prime, you need to show that
$$12! \equiv -1 \pmod {13^2} \,.$$  
To make the computations faster, observe that 
$$12!= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot (2 \cdot 5)\cdot 11 \cdot (4 \cdot 3)$$
We split the product in three:
$$2 \cdot 3 \cdot 4  \cdot 7 =(2  \cdot 7)(3 \cdot 4)=(13+1)(13-1)\equiv -1 \pmod {13^2}  $$
Also
$$6  \cdot 11 \cdot 8 \cdot 2 \cdot 4 = (6 \cdot 11)(8 \cdot 8)=(5 \cdot 13+1)(5 \cdot 13-1) \equiv -1 \pmod{13^2}$$
and
$$ 5   \cdot 9 \cdot  \cdot 5  \cdot 3=25 \cdot 27=(2 \cdot 13-1)(2 \cdot 13+1) \equiv -1 \pmod {13^2} \,.$$
Multiplying we get 
$$12! \equiv -1 \pmod{13^2} \,.$$
