Finding the sum of digits of $9 \cdot 99 \cdot 9999 \cdot ... \cdot (10^{2^n}-1)$

Let us assume that $N$ is a number with $2^k-1$ decimal digits, whose last digit is $\geq 1$.
Let $S(N)$ be the sum of digits of $N$. Let us study the sum of digits of $$ N\cdot(10^{2^k}-1) = N\cdot 10^{2^k}- N = N\cdot 10^{2^k} - 10^{2^k} + (10^{2^k}-1-N)+1. $$ We have:

$$ S(N\cdot(10^{2^k}-1)) = S(N)-1+\left(9\cdot(2^k-1)-S(N)+9\right)+1 $$ and it is very interesting to notice that such sum does not depend on $S(N)$, but simply is $9\cdot 2^k$.
The number

$$ N = 9 \cdot 99 \cdot 9999 \cdots (10^{2^{k-1}}-1) $$ has $2^k-1$ decimal digits, the last of them being $1$ or $9$.
By induction it follows that $$ S\left(9\cdot 99\cdots (10^{2^k}-1)\right) = \color{red}{9\cdot 2^{k}}.$$