Answer :
Let's do that by induction :
For [tex]n=1[/tex], [tex]x^1-y^1[/tex] is obviously divisible by [tex]x-y[/tex]
If we assume the property holds at rank [tex]n[/tex], then [tex]x^{n+1}-y^{n+1}=x(x^n-y^n)+y^n(x-y)[/tex]. Since [tex]x^n-y^n[/tex] is divisible by [tex](x-y)[/tex], we have [tex]A[/tex] such that [tex]x^n-y^n=A(x-y)[/tex] hence [tex]x^{n+1}-y^{n+1}=(x-y)(Ax+y^n)[/tex].
Hence by induction for all [tex]n\ge1[/tex], [tex]x-y[/tex] divides [tex]x^n-y^n[/tex]
For [tex]n=1[/tex], [tex]x^1-y^1[/tex] is obviously divisible by [tex]x-y[/tex]
If we assume the property holds at rank [tex]n[/tex], then [tex]x^{n+1}-y^{n+1}=x(x^n-y^n)+y^n(x-y)[/tex]. Since [tex]x^n-y^n[/tex] is divisible by [tex](x-y)[/tex], we have [tex]A[/tex] such that [tex]x^n-y^n=A(x-y)[/tex] hence [tex]x^{n+1}-y^{n+1}=(x-y)(Ax+y^n)[/tex].
Hence by induction for all [tex]n\ge1[/tex], [tex]x-y[/tex] divides [tex]x^n-y^n[/tex]
