Sum of the digits of $99^{99}$?

$99^{99}$ is less than $10^{198}$, therefore has at most 198 digits. When you sum them up (first time), you get a number that is less than or equal to $9\times 198=1782$. When you sum up the digits of that number (second time), you get something less than or equal to $\max(9+9+9, 1+6+9+9)=27$. When you sum up the digits of this last number, you get something less than or equal to $\max(2+7, 1+9)=10$.

But the original number is a multiple of $9$, hence the last number has to be a multiple of $9$. Note that it cannot be zero, because only $0$ itself has a sum of digits equal to $0$ and you started from $99^{99}\neq 0$. Therefore it is $9$.

$$99^{99}<100^{100}=10^{200}$$ so $99^{99}$ has at most $200$ digits and its digit sum is at most $1800.$ The digit sum of the digit sum is at most $28$ (a $1$ and three $9$'s). The digit sum of the digit sum of the digit sum is at most $11$, and it must be divisible by $9$, so it can only be $9$.