Polynomial ,divides and Induction Proof? 
$\text{The polynomial } x-y \;\text{divides the polynomial}\; x^2-y^2 \text{ and } x^3-y^3 \text{because}\; x^2-y^2 = (x+y)(x-y) \text{ and } x^3-y^3=(x-y)(x^2+xy+y^2.) \; \text{for every natural number n       } \quad x-y \;\text{ divides  }\; x^n-y^n \text{ prove by induction.}$

What I am confused about in this example is when one makes $n=k$. Which variable is made equal to k.Where is $n$. Is it fair to assume that what the actual problem wants you to infer this. $x^n +y^n = x^n+y^n$
(i) Basis Step:  $x^n-y^n = x^n-y^n$ 
$x-y = x-y$
$0=0$
(ii) Inductive step :  $x^k-y^k = x^k-y^k$
WNTS: RHS: $n =k+1$ 
$x^{k+1} + y^{k+1}$
LHS: $x^{k+1} + y^{k+1} +x^k-y^k $
Is this a fair assumption to make I feel as if their is something fundamentally missing from my logic. Any suggestions would be good.

For the inductive step, suppose $x-y$ divides $x^n-y^n$ for some $n$. You have to prove that, under this hypothesis, $x-y$ divides $x^{n+1}-y^{n+1}$.
Hint:
Rewrite it as $$x^{n+1}-y^{n+1}=x(x^n-y^n)+xy^n - y^{n+1}$$
and make partial factorisations.
Some details:
By the induction hypothesis, $x^n-y^n=(x-y)q(x,y)$, so 
$$x(x^n-y^n)+xy^n - y^{n+1}=(x-y)q(x,y)+ (x-y)y^n=(x-y)\bigl(q(x,y)+y^n\bigr).$$
Actually, the induction step defines a recurrence relation for the quotient of $x^n-y^n$ by $x-y$, which allows to  prove the complete factorisation formula.
