Prove that x $(x^{n-1}-n{a}^{n-1})+a^n(n-1)$ is divisible by $(x-a{)}^2$ for all positive integers n greater than 1

Let $A_n=x(x^{n-1}-n{a}^{n-1})+a^n(n-1)$
For n = 2, $A_2=x(x-2a)+a^2$
$=x^2-2ax{a}^2=(x-a{)}^2.$
Thus $A_2$ is divisible by $(x-a{)}^2.$
Now assume that $A_m$ is divisible by $(x-a{)}^2$ for m $\ge$ 2, that is, assume
$A_m=x(x^{m-1}-m{a}^{m-1})+a^m(m-1)$
$=(x-a{)}^2$ f (x)
so that $x^m=mx{a}^{m-1}-a^m(m-1)+(x-a{)}^{2}f(x)$
We then have
$A_{m-1}=x[x^m-(m+1\left)a^m\right]+a^{m+1}m$
$=x.x^m-(m+1)x{a}^m+m{a}^{m,+1}$
$=x[mx{a}^{m-1}-a^m(m-1)+(x-a{)}^{2}f\left(x\right)]$
$-(m+1)x{a}^m+m{a}^{m+1}$ by
$=m{a}^{m-1}\left[\right(x^2-2xa+a^2)+x(x-a{)}^{2}f\left(x\right)]$
$=m{a}^{m-1}\left(\right(x-a{)}^2+x(x-a{)}^{2}f(x)$
$=(x-a{)}^2[m{a}^{m-1}+xf\left(x\right)],$
After simplification.
This shows that $A_{m+1}$ is divisible by $(x-a{)}^2.$
Hence by induction, $A_n$ is divisible by $(x-a{)}^2$ for all positive integers n > 1.