Someone had posted a question on this site as to what would be sum of digits of $999999999999^3$ (twelve $9s$ ) equal to?
I did some computation and found the pattern that sum of digits of $9^3 = 18$, $99^3 = 36$, $999^3 = 54$ and so on. So, I had replied that sum of digits of $999999999999^3 = 12 \cdot 18 = 216$.
Can anybody help me prove this, that sum of digits of $(\underbrace{999\dots9}_{n\text{ times}})^3 = 18n.$