Show that (x-1)^2 is a factor of.......?

Question

Show that $(x-1)^2$ is a factor of $x^n-nx+n-1$

Here's my approach:- Letting $P(x)=x^n-nx+n-1$ I tried putting $P(1)=0$ but it would only prove for $,(x-1)$

Here's what I can think of, $x^n-1-nx+n=(x^n-1)-(nx-n)=(x-1)(x^{n-1}+x^{n-2}+x^{n-3}................+x+1)-n(x-1)$

Because $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^{n+1}................+xy^{n-2}+y^{n-1})$

$=(x-1)[x^{n-1}+x^{n-2}+x^{n-3}................+x+1-n]$

But I am not able to proceed any further neither I am able to get $(x-1)^2$ as common factor.

Answer 1

$P(x)=x^n-nx+n-1=(x^n-1)-(nx-n)$

$=(x^n-1)-n(x-1)=(x-1)(x^{n-1}+x^{n-2}+...+x+1-n)$

$P(1)=((1)-1)((1)^{n-1}+(1)^{n-2}+...+(1)+1-n)=(0)(0)$

Answer 2

Alternatively, experimenting with a few cases suggests the factorization $$x^n-nx+n-1=(x-1)^2(x^{n-2}+2x^{n-3}+3x^{n-4}+\cdots +(n-2)x+(n-1))$$ which is readily verified.