Try to use a function of $n$ to represent the product
$$
9 \times 99 \times 9999 \times \cdots \times\left(10^{2^{n}}-1\right)
$$

the sum of the digits in decimal notation.
Let $A_{n}=9 \times 99 \times 9999 \times \cdots \times\left(10^{2^{*}}-1\right)$. Clearly, the last digit of $A_{n}$ is not 0.
$A_{1}$ is a 3-digit number, and assuming $A_{n-1}$ is a $2^{n}-1$ digit number, then
$$
\begin{aligned}
A_{n} & =A_{n-1}\left(10^{2^{n}}-1\right) \\
& =10^{2^{n}} \cdot A_{n-1}-A_{n-1} \\
& =10^{2^{n}} \cdot\left(A_{n-1}-1\right)+\left(10^{2^{n}}-A_{n-1}\right)
\end{aligned}
$$

is a $2^{n}+\left(2^{n}-1\right)=2^{n+1}-1$ digit number.
Moreover, from (1) we can also see that the sum of the digits of $A_{n}$
$$
\begin{aligned}
S\left(A_{n}\right)= & S\left(A_{n-1}-1\right)+S\left(10^{2^{n}}-A_{n-1}\right)=S\left(A_{n-1}\right) \\
& -1+9 \times 2^{n}-S\left(A_{n-1}\right)+1=9 \times 2^{n} .
\end{aligned}
$$