TITLE: If $a$ is not a power of $10$, when $\lim\limits_{n\to\infty} S(a^n)=\infty$? ($s(m)$ is the sum of the digits of $m$.)
QUESTION [5 upvotes]: If $a$ is not a power of $10$, when $$\lim\limits_{n\to\infty} S(a^n)=\infty?$$ ($s(m)$ is the sum of the digits of $m$.)
(I don't have any ideas to prove this.)

REPLY [3 votes]: Here is a proof for the weaker claim that
$$\lim\sup s(a^n)=\infty,$$
if $a\geq2$ is an integer coprime to $10$.

Let $a\geq2$ be an integer that is coprime to $10$. For each integer $n\geq1$ define $l(n)$ to be the smallest integer such that $a^n<10^{l(n)}$. Then by Fermat's little theorem we have
$$a^{4\cdot10^{l(n)-1}}\equiv1\pmod{10^{l(n)}},$$
and hence also
$$a^n\equiv a^{n+4\cdot10^{l(n)-1}}\pmod{10^{l(n)}},$$
which means that
$$a^{n+4\cdot10^{l(n)-1}}=q\cdot10^{l(n)}+a^n,$$
for some integer $q$, and clearly $q$ is positive. It follows that
$$s(a^{n+4\cdot10^{l(n)-1}})>s(a^n).$$
So for the strictly increasing sequence $b_1,b_2,\ldots$ defined by $b_1=1$ and
$$b_{n+1}:=b_n+4\cdot10^{l(b_n)-1},$$
we have, for all $n\geq1$, the strict inequality
$$s(a^{b_{n+1}})>s(a^{b_n}).$$
This shows that
$$\lim\sup s(a^n)=\infty.$$