The hint states :
$a^{n+1} -a(a-1)n -a$
$= (a-1)(a^n+a^{n-1}+\ldots+a -n) $
$ =(a-1)((a^n-1)+(a^{n-1}-1)+\ldots+(a-1))$
$=(a-1)^2(a^{n-1}+2a^{n-2}+\ldots+n)$
But, I am confused at each step, particularly the first step in the division of powers of $a$, due to taking common factor as $(a-1)$. Similarly, in the last step, unable to divide to get second common factor again as $(a-1)$. Can I be guided at each step.
Edit Based on answers and comments, have progressed till:
I have computed incompletely, as follows:
$a^{n+1} -a(a-1)n -a = a((a^n-1) - (a-1)n) $ $= (a-1)(a(a^{n-1} + a^{n-2} + a^{n-3}+\ldots+a+1) -n)$ $=(a-1)((a^{n} + a^{n-1} + a^{n-2}+\ldots+a^2+a) -n)$.
For the last step, have to take another factor of $(a-1)$ again.
Taking cue from the earlier factorization, $(a-1)^2[{(a^{n-1} + a^{n-2} + a^{n-3}+\ldots+a+1)} + {(a^{n-2} + a^{n-3}+\ldots+a+1)} +{(a^{n-3} + a^{n-4} + a^{n-3}+\ldots+a+1)}+{(a^{n-4} + a^{n-5} + a^{n-6}+\ldots+a+1)}+\ldots+{a+1}+{1}]$.
So, each term apart from the first & last, i.e. $a^{n-1}, 1$ appears more than once with coefficients same as in Pascal's triangle(combinatorial expansion).
Also, there are $n-1$ terms' expansion apart from the last term $1$, so there are $(n-1)$ times extra occurrences of $1$, totaling to $(n-1)+1=n$..
Hence, proved.
Let $p(x)$ be the polynomial:
$p(x):=x^{n+1} + -x(x-1)n -x$ ,$ n \in \mathbb{Z^+}.$
$p(1) = 0$, hence
$p(x) = q(x)(x-1)$, where $q(x)$ is a polynomial of degree $n$.
Hence $(x-1)$ divides $p(x).$