Prove that $s(N(10^{n}-1)) = s(10^{n}-1)$ 
Let $N$ be a positive integer, and let $n$ be the number of digits in $N$ in decimal representation. Also let $s(n)$ denote the sum of the digits of $n$.  Prove that $s(N(10^{n}-1)) = s(10^{n}-1)$.

For example, we have $43 \cdot 99 = 4257$ and $s(4257) = 18 = s(99)$. How do we prove this result in general, and does it hold for all $m \geq n$?

Here I am assuming that the last digit $a_0 \neq 0$, if that's not the case then go for the last non-zero digit in your number $N$ and modify the proof accordingly.
Let 
$$N=a_{k-1}10^{k-1}+\dotsb +a_0, \qquad \text{ where } a_i \in \{0,1,\ldots ,9\} \text{ and } a_0 \neq 0$$
Then
\begin{align*}
N(10^k-1) & =[a_{k-1}10^{k-1}+\dotsb +a_0](10^k-1)\\
& =a_{k-1}10^{2k-1}+\dotsb +a_010^{k}-(a_{k-1}10^{k-1}+\dotsb +a_0)\\
& =a_{k-1}10^{2k-1}+\dotsb +(a_0-1)10^{k}+((9-a_{k-1})10^{k-1}+(9-a_{k-2})10^{k-2}+\dotsb +(10-a_0)).\\
\end{align*}
Thus
\begin{align*}
s(N(10^k-1)) &= \overbrace{9+9+\dotsb+9}^{k}\\
&=s(10^k-1). 
\end{align*}
