10.55 Find the digital sum of the product (as a function of $n$)
$$
9 \cdot 99 \cdot 9999 \cdot \cdots \cdot\left(10^{2^{n}}-1\right) \text {. }
$$

where the number of digits of each factor is twice the number of digits of the preceding factor.
(21st United States of America Mathematical Olympiad, 1992)
[Solution] Let $A_{n}=9 \cdot 99 \cdot 9999 \cdots \cdots \cdot\left(10^{2^{n}}-1\right)$.
First, we use mathematical induction to prove: $A_{n}$ is a $2^{n+1}-1$ digit number.
(1) When $n=0$, $A_{0}=9$ is a one-digit number, and $2^{0+1}-1=1$, so the conclusion holds for $n=0$.
(2) Assume $A_{k}$ is a $2^{k+1}-1$ digit number.
Then, when $n=k+1$,
$$
A_{k+1}=A_{k} \cdot\left(10^{2^{k+1}}-1\right)=A_{k} \cdot 10^{2^{k+1}}-A_{k} .
$$

Since $A_{k}$ is a $2^{k+1}-1$ digit number, from the above equation, $A_{k+1}$ is $\left(2^{k+1}-1\right)+2^{k+1}=$ $2^{(k+1)+1}-1$ digit number.
Thus, the conclusion holds for $n=k+1$.
Therefore, for all natural numbers $n$, $A_{n}$ is always a $2^{n+1}-1$ digit number.
Let the sum of the digits of $A_{n}$ be $S_{n}$. Since
$$
\begin{aligned}
A_{n} & =A_{n-1}\left(10^{2^{n}}-1\right) \\
& =\left(A_{n-1}-1\right) \cdot 10^{2^{n}}+10^{2^{n}}-A_{n-1} .
\end{aligned}
$$

Since $A_{n-1}$ is a $2^{n}-1$ digit number, we can set
$$
A_{n-1}=\overline{a_{1} a_{2} \cdots a_{m}}, \quad m=2^{n}-1, \quad a_{m} \neq 0 .
$$

Thus, we have
$$
\begin{aligned}
S_{n}= & \left(a_{1}+a_{2}+\cdots+a_{m}-1\right)+\left[9+\left(9-a_{1}\right)+\left(9-a_{2}\right)+\cdots+\right. \\
& \left.\left(9-a_{m-1}\right)+\left(10-a_{m}\right)\right] \\
= & 9 \cdot 2^{n} .
\end{aligned}
$$