A recent question asked about the sum of sum of sum of digits of $4444^{4444}$. The solution there works mainly because the number chosen is small enough for the sum of sum of sum to be equal to the repeated sum: i.e. if we sum digits further, the result does not change. Since finding repeated sums of digits is just a matter of elementary number theory, this solves the problem.
It seems the following question might be much harder: what is the sum of sum of digits of $$4444^{4444^{4444}}?$$ In other words, let $f:\Bbb N_0\to\Bbb N_0$ be the function defined by $f(n)=\textrm{sum of decimal digits of }n$.
What is the value of $f\left(f\left(4444^{4444^{4444}}\right)\right)$?
In this question, we have not yet reached a single-digit number, which at least seems to make it much harder.
Some estimates: the number of decimal digits of $4444^{4444^{4444}}$ is equal to $$\left\lfloor\log_{10}4444^{4444^{4444}}\right\rfloor+1,$$ which implies $$f\left(4444^{4444^{4444}}\right)\le9\left(\log_{10}4444^{4444^{4444}}+1\right).$$
Next, the number of digits of this last number is at most $$\left\lfloor\log_{10}\left(9\left(\log_{10}4444^{4444^{4444}}+1\right)\right)\right\rfloor+1,$$ which is $16213$, according to Wolfram|Alpha. Therefore, $$f\left(f\left(4444^{4444^{4444}}\right)\right)\leq9\cdot16213=145917.$$
So the number we are looking for has at most $6$ digits. This makes it very feasible to express in decimal notation, but possibly hard to find.
We might be further interested in numbers like $$f\left(f\left(f\left(4444^{4444^{4444^{4444}}}\right)\right)\right),$$ so a related question would be:
Is there any hope for a general method of evaluating such functions or is the behaviour of the $k$-fold composition $f^k$ completely chaotic?
