Write down the sum of sum of sum of digits of $4444^{4444}$

The approach is to use the fact that $4444 \equiv 7 \pmod 9,$ so that $4444^3 \equiv 1 \pmod 9,$ and then get $4444^{4444} \equiv 7 \pmod 9$.

Then use the fact that for any integer $N$, the sum of the digits of N is equivalent to $N \pmod 9$.

Finally use logs to base 10 to get a limit on the size of $A$, hence $B$ etc.

The answer is 7, if I remember correctly.

It's well-known that, because $10 \equiv 1 \bmod 9$, and therefore $10^k = 1 \bmod 9$ for all $k>0$, we have that the sum of digits of $n$, $S(n) \equiv n \bmod 9$.

So what is $4444^{4444} \bmod 9$? The above equivalence gives us that $4444 \equiv 7 \bmod 9$.

Now we need the order of $7$ modulo $9$, the smallest $s$ such that $7^s \equiv 1 \bmod 9$. This is easy to find by examination: $7^2 \equiv 4 \bmod 9$ and so $7^3 \equiv 28 \equiv 1 \bmod 9$.

So $4444^{4444} \equiv 7^{4444} \equiv 7 \cdot (7^3)^{1481} \equiv 7 \ \bmod 9$, and the same equivalence is true for $B,C$ and $D$.

Now roughly how big is $A$? Since $\log_{10}4444 \approx 3.64777$, we know that $\log_{10}A \approx 16210.7$, that is $A$ has $16211$ digits. This gives us that $B \le 16211\times 9 = 145899$. So if $B>100000$, the first two digits sum to no more than $5$, which means that $C \le 45$ (that maximum being when $B=99999$).

Finally we can use our knowledge that $C \equiv 7 \bmod 9$ and observe that $C$ must be one of the values $\{7,16,25,34,43\}$ and thus that $S(C) = \fbox{D = 7}$.

Additional thoughts:

We get exactly the same answer, $D=7$, for $A=55555^{55555}$ . Impressively, a slight variation on the same process also works to get $D=9$ for the case $A=999999999^{999999999}$ (nine nines).