Number of square numbers in the sequence?

Consider the $n$ -digit number $x_n=\underbrace{666\cdots6}_{\times n-1}7$ (which contains $n-1$ copies of $6$ ). Then $x_n = 2 \times \underbrace{333\cdots3}_{\times n} + 1 = \tfrac23(10^n-1) + 1 = \tfrac13(2 \times 10^n + 1)$ and so $x_n^2 \; = \; \tfrac19\big[4 \times 10^{2n} + 4 \times 10^n + 1\big] \; = \; \tfrac19\big[4(10^{2n}-1) + 4(10^n-1) + 9\big] \; = \; \tfrac49(10^{2n}-1) + \tfrac49(10^n-1) + 1$ Thus $x_n^2$ is the number obtained by adding $\underbrace{444\cdots4}_{\times 2n}$ (with $2n$ copies of $4$ ) to $\underbrace{444\cdots4}_{\times n}$ ( $n$ copies of $4$ this time), and then adding $1$ . In other words $x_n^2 \; =\; \underbrace{44\cdots4}_{\times n}\underbrace{88\cdots8}_{\times n-1}9$ which is the $n$ th number in this list.

This list consists of the numbers $x_1^2\,,\,x_2^2\,,\,\ldots\,,\,x_{100}^2$ , and so there are $\boxed{100}$ squares in the list.

Here's an indirect attempt at the bonus.

Consider the number $x=\frac{2}{3}+\frac{1}{3000}$ . Let's find $x^2$ . Expanding, we have $x^2=\frac{4}{9}+\frac{4}{9000}+\frac{1}{9000000}$ .

Now, consider the decimal expansions of these numbers.

$x=0.667$ (note that the $\frac{1}{3000}$ term "rounds" the fraction off at $3$ decimal places)

$\frac{4}{9}=0.444444444\ldots$

$\frac{4}{9000}=0.000444444\ldots$

$\frac{1}{9000000}=0.000000111\ldots$

Summing these, $x^2=0.444889$ , which proves the third number in the sequence is a square.

Exactly the same argument holds if we put $x=\frac{2}{3}+\frac{1}{3\cdot 10^n}$ , and proves the $n^{th}$ number in the sequence is a square for all $n$ .

(In particular, all $\boxed{100}$ of the numbers listed in the problem are squares.)

Bonus: Can you prove all numbers in that sequence (unlimited) are square numbers? In other word, prove 444...4488...889 (n fours, n-1 eights and 1 nine) is always a square number with n is a whole number.