The act of adding the digits of a number is called addichiffre. For example, when we addichiffre 124, we get $1+2+4=7$.

What do we get when we addichiffre $1998^{1998}$, then addichiffre the result obtained, and so on, three times in a row?
1998 and $1998^{1998}$ are multiples of 9, and therefore their digit sums are also multiples of 9. By induction, all digit sums obtained by repeated digit sum operations starting from $1998^{1998}$ are multiples of 9.

Moreover:

$$
1998^{1998} \leq 2000^{1998}=(2 \cdot 1000)^{1998}=2^{1998} \cdot 10^{3 \cdot 1998}=2^{1998} \cdot 10^{5994}
$$

and we know that $2^{9}=512<10^{3}$ and therefore $2^{1998}=2^{9 \cdot 222}<10^{3 \cdot 222}=10^{666}$ and thus $1998^{1998} < 10^{5994+666}=10^{6660}$. Therefore, the digit sum of $1998^{1998}$ is less than $9 \cdot 6660 = 59940$. The digit sum of a number less than 59940 is necessarily less than $5 + 9 \cdot 4 = 41$. The digit sum of a number less than 41 is less than $4 + 9 = 13$ and since this digit sum is a multiple of 9 and non-zero, this number must be 9.