How to prove $x^p\equiv x\mod p$ I have already proved that $(x+y)^p\equiv x^p+y^p\mod p$. 

And if I take $y=1$, how do I prove the congruency $x^p\equiv x\mod p$? (I should use induction to $x$ but I dont know how to do that in this case)
$$(x+1)^p\equiv x^p+1\equiv x+1\pmod p$$
Using Euler's Theorem, we have that  $x^p\equiv x\cdot x^{p-1} \equiv x\cdot x^{\phi(p)}\equiv x\cdot 1 \equiv x (mod p)$.