Prove $a^p\equiv a\pmod p$ Prove $a^p\equiv a\pmod p$, $p$ is a prime number.
Well, apart from writing this as $ a^p \pmod p = a \pmod p$ I don't know how to continue from here.
 A: I'm assuming $p$ is prime.  If $p\mid a$, then $a^p\equiv 0 \equiv a \pmod{p}$ and we're done.  If $p\nmid a$ then notice that both $\{1,2,\ldots,p-1\}$ and $\{a,2a,3a,\ldots,(p-1)a\}$ are reduced residue systems modulo $p$, and so the numbers in one set have to be congruent to the numbers in the other set, in some order.  Therefore the products of both sets are congruent modulo $p$.
$$1\cdot 2\cdot 3\cdots (p-1) \equiv a(2a)(3a)\cdots (p-1)a \pmod{p}.$$
Rearrange a bit:
$$(p-1)! \equiv (p-1)!a^{p-1} \pmod{p}.$$
Since $(p-1)!$ is relatively prime to $p$, we can divide it out:
$$1\equiv a^{p-1} \pmod{p}.$$
Now multiply through by $a$ and we're done. 
A: First of all, rewrite the theorem : let $p$ be a prime number $\forall a \in \mathbb{N}, a^p \equiv a [p]$
1:
For all we show $\forall k, 0<k<p, p|\binom{p}{k}$.
We know $k\binom{p}{k}=p\binom{p-1}{k-1}$ (you can show that easily), and $0<k<p$ so $\gcd(k,p)=1$ and by the gauss theorem $p|\binom{p}{k}$.
2: By induction on $a$ :
For $a=0$ it is ok.
Let assume that it is true $\forall a\in \{1,..p-2\}$ by the binomial formula we have :
$(a+1)^p=n^p+\sum_{k=1}^{p-1}\binom{p}{k}n^k+1\equiv a+1 [p]$ because $\forall k \in \{1,..p-1\},\binom{p}{k}\equiv 0 [p]$
