<s>[INST] Why does this pattern work: $1 \cdot{1} = 1, 11 \cdot{11} = 121, 111 \cdot{111} = 12321\ldots$ I have recently learned about pattern that goes like this:
$$\begin{align}
1^2 &= 1\\
11^2 &= 121\\
111^2 &= 12,321\\
1,111^2 &= 1,234,321\\
11,111^2 &= 123,454,321.
\end{align}$$
It is a very cool pattern, but after a bit it stops:
$$1,111,111,111^2 = 1,234,567,900,987,654,321$$
My main question:

Why does this pattern work?

A side question that's less important:

Is there an algebraic equation to describe this pattern?

 [/INST] Since OP is asking for an algebraic explanation, I will try a different approach, possibly trickier than the other answer which is elegant but not formal.
We consider the polynomial $(1+x+x^2+\dots +x^n)^2$: by expanding the square we get
$$\begin{array}{|c|c|c|c|c|c|c|}
\hline
1 & x & x^2 &\cdots& x^{n-1}& x^n\\ 
\hline
x & x^2 & x^3 &\cdots& x^{n}& x^{n+1}\\ 
\hline
x^2 & x^3 & x^4 &\cdots& x^{n+1}& x^{n+2}\\ 
\hline
\vdots & \vdots & \vdots &\vdots& \vdots& \vdots\\ 
\hline
x^{n-1} & x^{n} & x^{n+1} &\cdots& x^{2n-2}& x^{2n-1}\\ 
\hline
x^n & x^{n+1} & x^{n+2} &\cdots& x^{2n-1}& x^{2n}\\ 
\hline
\end{array}$$
that is, for $n\geq 0$, and for $0\leq j\leq 2n$,
$$\begin{align}
(1+x+x^2+\dots +x^n)^2&=\sum_{j=0}^{n}\sum_{k=0}^{j} x^k\cdot x^{j-k}
+\sum_{j=n+1}^{2n}\sum_{k=j-n}^{n} x^k\cdot x^{j-k}\\
&=
\sum_{j=0}^{n}x^j\sum_{k=0}^{j} 1
+\sum_{j=n+1}^{2n}x^j\sum_{k=j-n}^{n} 1\\
&=\sum_{j=0}^{n}(j+1)x^j
+\sum_{j=n+1}^{2n}(2n-j+1)x^j.
\end{align}$$
In our case, we take $x=10$. The pattern works as soon as
$j+1\leq 9$ for $j=0,\dots,n$ AND $2n-j+1\leq 9$ for $j=n+1,\dots,2n$ where $9$ is the largest decimal digit, that is for $n\leq 8$.
For $n=8$ and $x=10$ we have that
$$(1+x+x^2+\dots +x^n)^2=111111111^2=12345678987654321$$
whereas for $n=9$ we get
$$(1+x+x^2+\dots +x^n)^2=1111111111^2=1234567900987654321.$$
</s>