How to expand $(x^{n-1}+\cdots+x+1)^2$ (nicely) sorry if this is a basic question but I am trying to show the following expansion holds over $\mathbb{Z}$:
$(x^{n-1}+\cdots+x+1)^2=x^{2n-2}+2x^{2n-3}+\cdots+(n-1)x^n+nx^{n-1}+(n-1)x^{n-2}+\cdots+2x+1$.
Now I can show this in by sheer brute force, but it wasn't nice and certainly wasn't pretty. So I am just wondering if there are any snazzy ways to show this? If it helps, I am assuming $x^m=1$ for some $m>n-1$.

HINT.-Why not using $x^{n-1}+\cdots+x+1=\dfrac{x^n-1}{x-1}$ and dividing
$x^{2n}-2x^n+1$ by $x^2-2x+1$?
You will easily find successively the coefficients $$1,2,3,4,\cdots,(n-2),(n-1) ,n,(n-1),(n-2)\cdots,4,3,2,1$$ with a symmetry like the binomial coefficients.
