For any natural number $n$, we denote $\mathbf{S}(n)$ as the sum of the digits of $n$. Calculate $\mathbf{S}^{5}\left(2018^{2018^{2018}}\right)$.
It is clear that $A \equiv B \equiv C \equiv D \equiv E(\bmod 9)$. Since 2 has order 6 modulo 9, we then find

$$
2018^{2018} \equiv 2^{2018} \equiv 0 \quad(\bmod 2) \text { and } 2018^{2018} \equiv 2^{2018} \equiv(-1)^{2018} \equiv 1 \quad(\bmod 3)
$$

so $2018^{2018} \equiv 4(\bmod 6)$, which means that

$$
E \equiv 2018^{2018^{2018}} \equiv 2^{2018^{2018}} \equiv 2^{4} \equiv 7 \quad(\bmod 9)
$$

On the other hand, $A<10^{4 \times 2018^{2018}}$ so $B<9 \times 4 \times 2018^{2018}<10^{2+4 \times 2018}$, then $C<9 \times 2+9 \times 4 \times 2018<73000$, so $D<6+4 \times 9=42$. Consequently, $E<3+9=12$ and, since $E \equiv 7$ $(\bmod 9)$, we deduce that $E=7$.