Special palindromic pattern of $(\frac{10^n-1}{9})^2$

Write out several values for $10^n-1$. You'll see $n=1 \to 9$, $n=2 \to 99$, $\cdots, n \to 99\underset{n}{\underbrace{\,\ldots\,}}99 $ and

$$\frac{99\overset{n}{\overbrace \ldots}99}{9} = 11\underset{n}{\underbrace{\,\ldots\,}}11 $$

and the squares of $11\ldots11$ form this pattern you've noticed,

\begin{align*} 1^2 &= 1 \\ 11^2 &= 121 \\ 111^2 &= 12321 \\ 1111^2 &= 1234321 \\ &\ \vdots \\ 111\ 111\ 111^2 &= 12\ 345\ 678\ 987\ 654\ 321 \\ \end{align*}

You're squaring $\frac{10^{n}-1}{10-1}=10^{n-1}\sum_{k=0}^{n-1}(\frac{1}{10})^k$. To square $\sum_k x^k$ note each $x^p$ with $0\le p\le 2n-2$ can be achieved in one of $p+1$ ways for $p\le n$ and $2n-p-1$ ways otherwise. (You can think of it as a binomial-theorem truncation of $(1+x)^{-2}$, though the high-power coefficients are a little different because some ways you could try to achieve them require a factor beyond $x^{n-1}$.) For $n\le 10$, all coefficients are single-digit numbers, giving the pattern you observed. You may wish to use similar logic to see, for example, what happens with cubes.

\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.