Can somebody simply explain Wilson's theorem (for a 13 year old) I am Rohan Kapur. This is my first time posting on the Mathematics site, although I am quite active on StackOverflow, the programming site. 
I am doing a Islamic Maths assignment at the moment for Humanities, and I came across the historical fact that Ibn al-Haytham proved Wilson’s theorem. I have seen on this wiki page:
Wilson's theorem Wiki
a natural number $n > 1$ is a prime number if and only if: ￼
$(n-1)! \equiv -1 \pmod n$
So I know what $(n-1)!$ means, its a factorial. But what does $\equiv$ with three lines there $(\mathrm{mod}\ n)$ mean. How does that mean its a prime number? $5$ is a prime number, ok. 
$5-1!$ is $4 \cdot 3 \cdot 2\cdot 1$
that equals $24$
and then $24$, $\equiv$ with three lines again, $(\mathrm{mod}\ n)$.
What does this mean?
Been looking for a simple answer, but  can't find one...
UPDATE
Yes, I know what modular arithmetic is. It is a system where a number wraps around after a certain value in a loop. Like a clock time for example, but what does this mean in proving that the number is a prime. 
 A: I don't have time to answer your question fully right now, because I'm about to go to lunch.  But why don't you try, for a start, showing that for any composite $n$ (except $n=4$), $(n-1)!$ is divisible by $n$.  (This is the same as saying $(n-1)! \equiv 0 \pmod{n}$.)  This, along with the observation that $(4-1)!\equiv 2 \pmod{4}$, gives you one half of the proof.
The other half is harder, and uses some things which I expect you don't know yet.  But let me know whether you can get the first half for a start.  I'll be back in about half an hour.
A: Wilson's Theorem, written without congruences, says that $n$ is prime if and only if $(n-1)!+1$ is a multiple of $n$. 
Let's see that it gives us the right answers for $n=4$ and $n=5$. 
$n=4$: $(n-1)!+1$ is $3!+1$ which is 7 which is not a multiple of 4, and 4 is not prime - good! 
$n=5$: $(n-1)!+!$ is $4!+1$ which is 25 which is a multiple of 5, and 5 is prime - good! 
So, it works for $n=4$ and $n=5$. 
Now instead of proving it works for every prime, I'll show you why it works for 11, and claim the same idea always works. 
For $n=11$, we have to look at $10!$, which is $(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)$. I'll rearrange that as $(1)((2)(6))((3)(4))((5)(9))((7)(8))(10)$, which is $(1)(12)(12)(45)(56)(10)$. Now notice that each of those numbers, $1,12,12,45,56$, is 1 more than a multiple of 11. That implies that when you multiply them together, you get 1 more than a multiple of 11. But the number I didn't account for, 10, is one less than a multiple of 11. So when you multiply that in, you get 1 less than a multiple of 11, as Wilson's Theorem asserts. 
Well, what happens for $n=11$ happens for every prime, although it takes a bit of work to establish it: you can always pair off all the numbers so that each pair multiplies to 1 more than a multiple of $n$, except the last number, $n-1$, which is one less than a multiple of $n$. So the whole $(n-1)!$ is one less than a multiple of $n$, as Wilson says. 
