1. 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 representation.
First, we prove the following lemma: Let the number of digits of the natural number $m$ be no more than $d$, and $M=\left(10^{*}-1\right) m(b \geqslant \geqslant i)$. Then $S(M)-S$ S.

Here $S(M)$ represents the sum of the digits of $M$.
Let $M=\left(10^{k}-1\right) m=p+q+r$, where
$$
\begin{array}{l}
p=10^{k}(m-1), q=10^{d}-m, \\
r=10^{k}-10^{d} .
\end{array}
$$

If the decimal representation of $m-1$ is
$$
m-1=\overline{a_{1} a_{2} \cdots a_{d}},
$$

then $p=\overline{a_{1} a_{2} \cdots a_{d} 00 \cdots 0}$ ( $k$ zeros ),
$$
\begin{aligned}
& q=\left(10^{d}-1\right)-(m-1) \\
& =99 \cdots 9-(m-1) . \\
& =99 \cdots 9(d \text{ 9s})-\overline{a_{1} a_{2} \cdots a_{d}} \\
& =\overline{b_{1} b_{2} \cdots b_{d}}, \\
a_{i}+b_{i}= & 9 \quad (i=1,2, \cdots, d) . \\
r= & \left(10^{k}-1\right)-\left(10^{d}-1\right) \\
& =99 \cdots 9(k \text{ 9s})-99 \cdots 9(d \text{ 9s}) \\
& =99 \cdots 900 \cdots 0(k-d \text{ 9s}, d \text{ 0s}) .
\end{aligned}
$$

Therefore, $M=\overline{a_{1} a_{2} \cdots a_{d} 00 \cdots 0}$ ( $k$ zeros ) $+\overline{b_{1} \cdots b_{d}}$

Thus, $S(M)=9 k$.
Let the given product be $M_{1}$, and
$$
m=9 \times 99 \times \cdots \times\left(10^{2^{n-1}}-1\right).
$$

Then $M_{1}=\left(10^{2^{n}}-1\right) m$.
Also, $m<10^{1+2+\cdots+2^{n-1}}=10^{2^{n-1}}$,
so the number of digits of $m \leqslant 2^{n}$. By the lemma,
$$
S\left(M_{1}\right)=9 \cdot 2^{n}.
$$