Is it true that for infinitely many values of $n$, the sum of digits of $2^n$ is greater than for $2^{n+1}$?

Expanding on Deven Ware's observation: assuming $S(2^n)$ is eventually nondecreasing, not only would it strictly increase, but we can give a non-trivial lower bound on its rate of growth.

Working mod $3$, for instance, note that if for some $m$, $S(2^m) \equiv 2 \pmod 3$, then $S(2^{m+1}) \equiv 1 \pmod 3$, so if $S(2^{m+1}) > S(2^m)$ then in fact $S(2^{m+1}) \ge S(2^m)+2$. We can easily see this leads a lower bound of $$S(2^n) \ge \tfrac32 n + O(1),$$ which is stronger than the $n+O(1)$ lower bound one gets from just being strictly increasing.

On the other hand, since $2^n$ has $\frac{\log 2}{\log 10}n + O(1)$ digits, we have an upper bound of $$S(2^n) \le 9\frac{\log 2}{\log 10}n + O(1) < 2.71n + O(1).$$ This doesn't contradict our lower bound of $S(2^n) \ge \tfrac32 n + O(1)$, but look at what happens when we work modulo $9$ instead:

The sequence $\{2^n\}$ cycles modulo $9$ as $1,2,4,8,7,5,1,\ldots$, and $\{S(2^n)\}$ follows exactly the same cycle. However, if we assume $S(2^n)$ is (eventually) nondecreasing, then starting from any sufficiently large $m$ such that $S(2^m) = 9k+1$, we have the chain of inequalities:

\begin{align} S(2^{m+1}) &\ge 9k+2, \\ S(2^{m+2}) &\ge 9k+4, \\ S(2^{m+3}) &\ge 9k+8, \\ S(2^{m+4}) &\ge 9k+16, \\ S(S^{m+5}) &\ge 9k+23, \\ S(S^{m+6}) &\ge 9k+28, \end{align}

which yields a lower bound of $S(2^n) \ge \frac{27}{6} n + O(1) = 4.5n + O(1)$. This does contradict the upper bound from the number of digits, so indeed $S(2^n)$ must decrease infinitely often.

What homework is this from? This is actually quite a pretty question, and I'm both pleased and impressed that it can be answered by such elementary methods.