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...
Pd. Indeed not "quite trivial"