Finding the sum of digits of $9 \cdot 99 \cdot 9999 \cdot ... \cdot (10^{2^n}-1)$ Find the sum of the digits (in terms of $n$) of $$9 \cdot 99 \cdot 9999 \cdot 99999999 \cdot ... 
 \cdot (10^{2^n} - 1)$$
Where every term has two times more digits than the previous term.
I tried to factorise as $$(10-1)(10^2-1)(10^4-1)...(10^{2^n}-1)$$
and they're all differences of two squares (except for the first term)
$$(10-1)(10-1)(10+1)(10^2-1)(10^2+1)(10^4-1)(10^4+1)...(10^{2^n-1}-1)(10^{2^n-1}+1)$$
but this doesn't seems to help that much. Also i tried with base 10 but it didn't work at all. Any ideas?
 A: 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
$$ S\left(9\cdot 99\cdots (10^{2^k}-1)\right) = \color{red}{9\cdot 2^{k}}.$$
A: OK, so the first couple of numbers and the sum of their digits are:
\begin{array}{ccc}
n & num(n) & sum(n)\\
\hline
0&9&9\\
1&891&18\\
2&8909109&36\\
3&890910891090891&72\\
\end{array}
So, this suggests that the sum of digits will be $9*2^n$, and if you look at the number, you'll understand why:
As we go from $n$ to $n+1$, the last digit from $sum(n)$ gets lowered by $1$, while digits get added after that that is the result of $10^{2^n}-num(n)$, and those digits add up to $sum(n)+1$, making $sum(n+1)=sum(n)-1+sum(n)+1=2*sum(n)$. So this is not a hard proof yet, but if you can prove these assertions, then an inductive proof will do the rest.
