I just want to give some other proof to your second question. Let $SS(m^n)=\lbrace m^n ; n\geq1\rbrace$. We show that this set is unbounded. First a Lemma:
$Lemma:$ Let $a$ be any arbitrary integer unless your trivial cases. For any arbitrary integer $b$, there is an integer $t$ such that $a^t$ begins with $b$.
Now suppose $Max$ $SS(m^n)=M$, and just look to this sequence of integer:
$\overline {M1}, \overline {M11}, \ldots$
this sequence is infinite and by previous lemma, for each of them you must have a suitable $t_i$, $i$ is the position of each number in the above sequence, that $m^n$ begins with $i$-th term in the sequences. Obviously, each of them has the digit sum greater than $M$ and it is a contradiction. This completes the proof.