Magic of 899.....9

We are interested in $(9 \times 10^n - 1)^2$ , which equals (when $n \ge 2$ ) $\begin{aligned} 81 \times 10^{2n} - 18 \times 10^n + 1 & = \; 81\overbrace{00 ... 0}^{\times n}\overbrace{00 ... 0}^{\times n} - 18\overbrace{00...0}^{\times n} + 1 \\ & = \; 80\overbrace{99...9}^{n-2}82 \overbrace{00...0}^{\times n} + 1 \; = \; 80\overbrace{99...9}^{\times n-2} \overbrace{00...0}^{\times n-1}1 \end{aligned}$ Thus there are $1 + n-1 = \boxed{n}$ $0$ s in $8\overbrace{99...9}^{\times n}$ , at least for $n \ge 2$ . Note that $89^2 = 7921$ has no zeros, so breaks this rule. You should specify that $n \ge 2$ here.

$899...9^2$

= $(900....0-1)^2$

= $(9*10^n-1)^2$

= $81*((10)^2)^n+1-2*9*(10)^n$

= $9*(10)^n[9*(10)^2-2]+1$

= $898*9*(10)^n+1$

= $8082*(10)^n+1$

∵ [ $8082*(10)^n$ has n+1 zeros]

∴ $8082*(10)^n+1$ has n number of zeros.