Probability That an $n$ Digit Integer's Digits Will Have a Product of Zero The Question
Let $d(n):\mathbb{N} \rightarrow \mathbb{N}$ be a function which given an input $n$ will map it to the product of its digits. Show that for sufficiently large $n$ the probability that $d(n) = 0$ is $1$.
My Work
There are $9 \times 10^{n-1}$ possible $n$ digit integers. There are $9^n$ possible $n$ digit numbers with no zeros. Therefore, there are $9 \times 10^{n-1} - 9^n$ possible $n$ digit integers with at least one zero. The probability of an $n$ digit natural number $k$ fulfilling the property $d(k) = 0$ is $\frac{(9 \times 10^{n-1} - 9^n)}{9\times 10^{n-1}}$
$\lim_{n \to \infty} \frac{(9 \times 10^{n-1} - 9^n)}{9\times 10^{n-1}} = 1$ (Via Wolfram Alpha)
My Question
This is a question I wondered about while working on a programming problem involving digit products. I don't know if the steps I took are correct or if what I'm trying to prove is even correct, so some verification would be nice. I'm also wondering if there is another solution to this problem, if it is correct?
 A: You solved this correctly. Another way to see this is any digit (except the leading digit) has a $\frac{9}{10}$ chance of not being  $0$. Then the probability that a random $n$ digit number's digits' product will not be zero is $(\frac{9}{10})^{n-1}$. We see as $n \to \infty$, this product goes to $0$.
A: Your proof is basically correct, but the statement you are trying to prove does not make sense.  What does it mean to say "for sufficiently large $n$ the probability that $d(n) = 0$ is $1$"?  For any particular value of $n$, $d(n)$ has a definite, specific value, and so either $d(n)=0$ or $d(n)\neq 0$ (it's not a matter of probability).  To talk about probabilities in a meaningful way, you need to say how you are choosing $n$ at random (say, you are choosing it randomly from a certain range of possible values, with all values being equally likely).  Moreover, the phrase "for sufficiently large $n$..." means "there exists a constant $C$ such that for all $n>C$...".  In this case, that would mean you are asserting that $d(n)=0$ for all $n>C$, which is not true (since you could take $n$ to be a number with more digits than $C$, none of which are $0$).
What you have proven (correctly) is that if you let $p_N$ be the probability that $d(n)=0$ if $n$ is a randomly chosen $N$-digit positive integer, then $\lim_{N\to\infty}p_N=1$.  That is, if you randomly choose $n$ among all $N$-digit numbers for some fixed $N$, the probability that $d(n)=0$ approaches $1$ as you make $N$ larger and larger.
