Solution:

Put $p(n)$ for the product of the digits of $n$. We show that, for sufficiently large $n$, a sequence starting below it cannot get past the "gap" from $10^{n}$ to $10^{n} + 10^{n - 1}$. For suppose $N$ is the last member of the sequence below the gap. Then $N$ has at most $n$ digits, so $p(N) \leq 9^{n}$. But for sufficiently large $n$ (in fact for $n \geq 21$) we have $9^{n} < 10^{n - 1}$. So $N + p(N) < 10^{n} + 10^{n - 1}$. But $N + p(N) > 10^{n}$ by assumption. Hence $N + p(N)$ is sure to have second digit (from the left) zero.
So all further terms of the sequence are the same. But for any $m$ there is certainly a gap above $m$, and, as shown, the sequence will not be able to get beyond it. So it is bounded.