Repeated digit multiplication: How often does it lead to zero?

Question

Take a number $n$ expressed in base $10$, and multiply its digits together to yield a new number. Repeat this process until a single digit is reached: \begin{eqnarray} 42433 &\to& 4 \cdot 2 \cdot 4 \cdot 3 \cdot 3 = 288 \\ 288 &\to& 2 \cdot 8 \cdot 8 = 128 \\ 128 &\to& 1 \cdot 2 \cdot 8 = 16 \\ 16 &\to& 6 \end{eqnarray}

Often a zero digit occurs along the way, which then maps to $0$ immediately: \begin{eqnarray} 31378 &\to& 3 \cdot 1 \cdot 3 \cdot 7 \cdot 8 = 504 \\ 504 &\to& 5 \cdot 0 \cdot 4 = 0 \end{eqnarray}

My question is:

Q. For a random number $n \in [1,N]$, what is the probability that this repeated digit multiplication leads to $0$?

It seems quite high, perhaps $90$% for large $N$, but I don't see a straightforward way to calculate it. It would also be of interest to know the distribution of the non-zero terminal digits. For example, even terminal digits occur more frequently than odd digits (for obvious reasons).

One can ask the same questions for numbers expressed in base $b$.

Answer 1

Given $\epsilon > 0$, for large enough $N$, more than $1-\epsilon$ of the integers in $[1,N]$ will have at least one $0$. That is, the fraction of $d$-digit integers with no $0$ is $(9/10)^{d-1}$, which goes to $0$ as $d \to \infty$.

Answer 2

The answer is greater than or equal to the proportion of numbers in $[1,N]$ that have the digit $0$, OR that have at least one $5$ and at least one even digit. As $N\to\infty$, this proportion approaches $1$.

The probability sought is greater than the proportion indicated above, because of numbers such as $2333$, whose digit product is $54$, which in turn produces a $0$ in the next step.