1992 USAMO 第 1 题

Find, as a function of $\, n, \,$ the sum of the digits of

$$9 \times 99 \times 9999 \times \cdots \times \left( 10^{2^n} - 1 \right),$$

where each factor has twice as many digits as the previous one.

求 $\, n, \,$ 的数字之和

$$9 \times 99 \times 9999 \times \cdots \times \left( 10^{2^n} - 1 \right),$$

其中每个因子的位数是前一个因子的两倍。