I have watched: Veritasium
The idea behind is that there exist value that satisfies: $n^2 = n$.
That value is its own square.
We can create this value by:
$$ 5^2 = 25 $$ $$ 25 ^ 2 = 625 $$ $$ 625^2 = 390625 $$ $$ 0625^2 = ... $$ $$ 90625^2 = 8212890625 $$
How we can prove that we can go creating this value for indefinitely ?
Here’s a rather abstract treatment of the phenomenon:
The $10$-adic numbers $\Bbb Z_{10}$ turn out to be the “ring-direct sum” of the $2$-adic numbers $\Bbb Z_2$ and the $5$-adic numbers $\Bbb Z_5$. We express this thus: $$ \Bbb Z_{10}\cong\Bbb Z_2\oplus\Bbb Z_5\,. $$ What’s on the right side of the isomorphism sign is the set of all pairs $(u,v)$ for which $u\in\Bbb Z_2$ and $v\in\Bbb Z_5$. Addition and multiplication are performed coefficientwise, with no interaction across the $\oplus$-sign. The zero-element is $(0,0)$ and the unity element is $(1,1)$. They behave correctly, that is, $(0,0)+(u,v)=(u,v)$ and $(1,1)\cdot(u,v)=(u,v)$, where I’m using the dot to denote multiplication.
But there are two other “idempotents”, i.e. objects $z$ for which $z\cdot z=z$, namely $\Bbb E_2=(1,0)$ and $\Bbb E_5=(0,1)$. You can get $\Bbb E_2$ by starting with a number $n$ that’s $\>\equiv1\pmod2$ but $\>\equiv0\pmod5$: The starting value $n=5$ will do nicely. Then, recursively, you square each value of $n$ to get the next value of $n$.
As you saw, the first few values of $n$ are $5,\,25,\,625,\,390625,\,152587890625,\,\cdots$
But after a few repetitions, you’re led to realize that at the $k$-th stage, you get stability in only the first $k$ digits: those to the left get changed. So, you might as well throw away all the digits to the left of the $k$-th digit. This makes your calculations more efficient. For instance, if you want the rightmost twelve digits of your magic number $\Bbb E_2$, you’ll find that they are $\>\cdots918212890625\,$. And since $E_2+E_5=(1,1)$, which has ten-adic expansion simply $1$, you can get the expansion of $\Bbb E_5$ by writing out $1 - E_2$ ten-adically, to get $\>\cdots081787109376\,$. You can check that this is truly also an idempotent, to twelve $10$-adic places, and finally you can check that the leftmost twelve digits of $918212890625\cdot081787109376$ all are zeros, since $\Bbb E_2\cdot\Bbb E_5=(1,0)\cdot(0,1)=(0,0)$.