Example 7 (17th IMO problem) When $4444^{444}$ is written as a decimal number, the sum of its digits is $A, B$ is the sum of the digits of $A$. Find the sum of the digits of $B$.


Solution: First, every positive decimal integer $n=a_{0} 10^{k}+a_{1} 10^{k-1}+\cdots+a_{k-1} \cdot 10+a_{k} \equiv a_{0}+a_{1}+\cdots+a_{k}(\bmod 9)$, which means every number is congruent to the sum of its digits modulo 9.

Secondly, since the number of digits in $4444^{4441}$ does not exceed $4 \times 4444=17776$, we have $A \leqslant 177760, B \leqslant 1+5 \times 9=46, B$'s sum of digits $C \leqslant 4+9=13$. And
$$
C \equiv B \equiv A \equiv 4444^{444} \equiv 7^{4444}=\left(7^{3}\right)^{1481} \times 7 \equiv 1^{1481} \times 7 \equiv 7(\bmod 9),
$$
thus $C=7$, which means the sum of the digits of $B$ is 7.