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?

 A: 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
$$(1+x+x^2+\dots +x^n)^2=1111111111^2=1234567900987654321.$$
A: Other answers have already provided some intuition behind the pattern. Here is some terminology:

*

*The numbers being squared, e.g. $111111$, are called repunits.


*The resulting products are called Demlo numbers: $\,1, 121, 12321, 1234321,\, \dots$

Repunits can be written as sums of powers of $10$, and from that perspective, base-$10$ repunits are finite geometric series. For example:
\begin{align}
111111 &= 1 + 10 + 100 + 1000 + 10000 + 100000\\
111111&=1 + 10 + 10^2 + 10^3 + 10^4 + 10^5\\\\
111111&=\displaystyle\frac{10^6-1}{10-1}\\\\
111111&=\displaystyle\frac{10^6-1}{9}
\end{align}
The last two lines above use the formula for the sum of a finite geometric series.
Squaring gives:
\begin{align}
111111^2 &= \left(\displaystyle\frac{10^6-1}{9}\right)^2\\
111111^2 &= \displaystyle\frac{10^{12} - 2\cdot 10^6 +1}{81}
\end{align}
In general:
$$\left(\,\underbrace{\,1111\,\dots\,1111\,}_{ n \text{ ones} } \,\right)^2 \, = \,\, \boxed{\displaystyle\frac{10^{2n} - 2\cdot 10^n +1}{81}\,}$$

Demlo numbers are sequence A002477 in the OEIS.
A: Just do the maths:
        1 1 1 1 1
      x 1 1 1 1 1
        ---------
        1 1 1 1 1
      1 1 1 1 1
    1 1 1 1 1
  1 1 1 1 1
1 1 1 1 1
-----------------
1 2 3 4 5 4 3 2 1

Of course the pattern only goes up to nine $1$'s, because after that you get eight hundred and ninety-ten in the billions column...
A: well, for 1^2 it isn't very hard to figure out, but here is the basic explanation. there is one of 11, 111, 1111, etc. in each place value until there have been as many place values as there are digits of the original number (1,111; 11,110; 111,100; and 1,111,000 for example). the amount of digits in the product is equal to (nx2)-1, where n is the original number. then, it is like a bell curve, with the numbers determined by how many ones were in that place value. it then breaks because there is no single digit bigger than 9, and thus it moves to the next place, and ruins the symmetry.
