Induction

I will use the term mod to mean the modulus with respect to (x-y)

For n=1, is it obvious that x-y is a factor of x^n - y^n

Assume that for a given n, (x^n - y^n) is evenly divisible by (x-y). Prove that [x^(n+1) - y^(n+1)] is also divisible by (x-y)

mod (x^n - y^n) = 0 [Given]
mod x^n = mod y^n [1]

Find the value of mod [x^(n+1) - y^(n+1)]
mod [x^(n+1) - y^(n+1)] =
mod x*x^n - mod y*y^n =
mod x * mod x^n - mod y * mod y^n
mod x^n * (mod x - mod y) [From 1]
mod x^n * mod (x-y), but mod (x-y)=0, so the entire expression equals zero therefore

if (x^n - y^n) is divisible by (x-y), then so is [x^(n+1) - y^(n+1)]

Therefore all values of x^n - y^n are divisible by (x-y)

You are too good man! Thanks!