Define a sequence as follows:

a_{1} = k

a_{n+1} = a_{n} + (the product of the digits of a_{n})

Is there a value of k for which this sequence is unbounded?

Note first that for a starting value k, if there is a term a_{n} in the
sequence that includes the digit 0, then all the subsequent terms of the
sequence are equal to a_{n}, and the sequence is bounded.

We can thefore prove the assertion by establishing that for any starting value k, there will be a term of the sequence that includes the digit 0.

So, proceeding by contradiction, we assume that for a given k, the sequence is unbounded, meaning that no term of the sequence includes the digit 0.

Pick an integer r such that (1) r >= 22 and (2) k < 10^{r}.
The sequence is monotone increasing, so let a_{m} be the
last term which is less than 10^{r}.
By (2), there must exist such a term.

What can we say about a_{m+1}?
First of all, by definition:

a_{m+1} >= 10^{r} (1)

Furthermore:

a_{m+1} = a_{m} + (the product of the digits of a_{m})

But a_{m} has at most r digits, each of which is at most 9, so:

a_{m+1} <= a_{m} + 9^{r} (2)

(10/9)^{r} >= (10/9)^{22} = 10.1546460985... > 10^{1}

Hence:

9^{r} < 10^{r-1}

Combining this with (1) and (2) above, we finally get:

10^{r} <= a_{m+1} < 10^{r} + 10^{r-1}

This implies that the second digit of a_{m+1} is 0, which is
the contradiction that establishes the result.