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 $m>n-1$.

A kind of graphical proof: Consider the case of $(1+x+x^2+x^3+x^4+x^5)^2$ expanded in a square array in this way:
$$\begin{array}{|l||l|l|l|l|l|}
\hline
&\color{red}{1}&\color{red}{x}&\color{red}{x^2}&\color{red}{x^3}&\color{red}{x^4}&\color{red}{x^5}\\ 
\hline
\color{red}{x^5}&x^5&x^6&x^7&x^8&x^9&x^{10}\\
\hline
\color{red}{x^4}&x^4&x^5&x^6&x^7&x^8&x^{9}\\
\hline
\color{red}{x^3}&x^3&x^4&x^5&x^6&x^7&x^{8}\\
\hline
\color{red}{x^2}&x^2&x^3&x^4&x^5&x^6&x^{7}\\
\hline
\color{red}{x}&x&x^2&x^3&x^4&x^5&x^{6}\\
\hline
\color{red}{1}&1&x&x^2&x^3&x^4&x^{5}\\
\hline
\end{array}$$
Terms $x^k$ with the same exponent $k$ are situated "in a natural way" on a same diagonal and the "population" of these diagonals linearly increase:
$$1, \ 2 x, \ 3 x^2, \ 4 x^3, \ 5 x^4, \ 6 x^5, \ 5 x^6, \cdots 2 x^9, \ 1x^{10},$$
with a maximum along the main diagonal, then linearly decrease...
Remark: one mimicks here the (discrete) convolution of a uniform distribution with itself giving a "tent" function, as it is called in signal processing, with an evident application : the law of the sum of two dies (here with faces numbered $0$ to $5$) with a maximal probability for result 5.
