Special palindromic pattern of $(\frac{10^n-1}{9})^2$ I noticed a repetitive, palindromic pattern when I've performed exponentiation with $10$. That is: $$(\frac{10^n-1}{9})^2 \text{ has a palindromic pattern, }\forall\:n\in [1;9]\:(n\in \mathbb{N})$$
We define $f(n)=(\frac{10^n-1}{9})^2$.

*

*$f(1) = \color{red}{1}$


*$f(2) = \color{red}{1}\color{blue}{2}\color{red}{1}$


*$f(3) = \color{red}{1}\color{blue}{2}\color{green}{3}\color{blue}{2}\color{red}{1}$
$$...$$

*

*$f(9) = \color{red}{1}\color{blue}{2}\color{green}{3}\color{pink}{4}\color{magenta}{5}6\color{grey}{7}\color{brown}{8}\color{orange}{9}\color{brown}{8}\color{grey}{7}6\color{magenta}{5}\color{pink}{4}\color{green}{3}\color{blue}{2}\color{red}{1}$*


I was wondering why this expression has such results. Not only they are palindromic, they are in fact using ascending consecutive digits until the sequence repeats, in descending order. Can anyone explain why the structure of these numbers is so strange?
Thanks in advance!

\begin{align*}
11 \times 11 &= \begin{matrix} 1 & 1 & \\ & 1 & 1 & + \\ 1 & 2 & 1 \end{matrix} \\ \\
111 \times 111 &= \begin{matrix} 1 & 1 & 1 & & \\ & 1 & 1 & 1 & \\ & & 1 & 1 & 1 & + \\ 1 & 2 & 3 & 2 & 1 \end{matrix} \\ \\
1111 \times 1111 &= \begin{matrix} 1 & 1 & 1 & 1 & & & \\ & 1 & 1 & 1 & 1 & & \\ & & 1 & 1 & 1 & 1 & \\ & & & 1 & 1 & 1 & 1 & + \\ 1 & 2 & 3 & 4 & 3 & 2 & 1 \end{matrix}
\end{align*}
etc. Due to the rotational symmetry of these parallograms, and the fact that, for $n \le 9$ there will be no carry-over in any column, you're going to get symmetric digits.
