At first sight this looks quite trivial, but let's see:

a^p = b^p (mod p) ==> a = b (mod p), since a^p = a (mod p) for any

integer a ==> a = b + kp ==> a^p = (b + kp)^p = b^p + (sum of

numbers all of which are divisible by p^2) ==> a^p = b^p (mod p^2)

The only non quite trivial thing here, imo, is why in that sum all the

numbers are divisible by p? Well, the first summand there is p(kp) and the

following ones are (binomial coefficient)*(kp)^i, with i >= 2, so...

Tonio

Pd. Indeed not "quite trivial"