Denote $p = 1/10$ to keep the formulas short.
Let $E$ be the expected number of digits the generator produces before we see the first $9$.
Then
$$ E = p \cdot 1 + (1-p) \cdot (1 + E) $$
because if we see a $9$ right away, the number of digits we needed was $1$, and if we don't, then we need one plus the number of additional digits.
This additional number of digits has the same expected value $E$ as the original, because the generator is "memoryless": seeing a non-$9$ doesn't make a $9$ on the next steps any more or less likely.
When we solve the equation, we get $E = 1/p$, which is $10$.
Then you want to compute the expected number of digits before seeing $n$ nines, not necessarily as a contiguous string (so something like $0 9 0 7 5 9 9 3 9$ would count as a success for $n = 4$).
Seeing $n$ nines is precisely the same as seeing one nine, then one nine after that, then one nine after that, and so on $n$ times.
The number of digits needed is the sum of the number of digits in these $n$ processes, and they are independent because the generator is memoryless.
The expected number of digits is $E + E + E + \cdots + E = E n = 10 n$ where there are $n$ copies of $E$ in the sum.
For any sequence $s$ of $n$ digits, the expected number of digits produced before $s$ occurs as a possibly non-contiguous subsequence is $10 n$, for the same reason.
For a contiguous sequence of $n$ nines, the computations are still pretty simple.
Let $E_n(k)$ be the expected number of additional digits the generator needs to produce before we see $n$ nines, given that we have just seen $k$ consecutive nines.
For example, if the generator has produced $4 5 9 2 2 9 9$, we'd be looking at $E_n(2)$.
Then $E_n(n) = 0$ because we don't need any more digits after $n$ nines, and for $0 \leq k < n$ we have
$$ E_n(k) = p(1 + E_n(k+1)) + (1-p)(1+E_n(0)) $$
because seeing a $9$ makes the just-seen sequence one digit longer, but seeing anything else resets it.
Now we have a finite system of linear equations in the $E_n(k)$ that we can solve.
By induction we can prove $E_n(0) = E_n(k) + \frac{p^{-k} - 1}{1 - p}$, and at $k = n$ this means $E_n(0) = \frac{p^{-n}-1}{1-p} = 10 (10^n - 1) / 9$.
Think of your numerator as $10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3$; that is, you have $10$ choices for the first digit, but then only $9$ for the second, $8$ for the third, and so on down to $3$ for the eighth digit. The decreasing number of choices reflects the fact that you are not repeating digits that have already been used. Alternatively, $10!/2!$ means you permute the ten digits in any of $10!$ ways, drop the final two digits, and group them in pairs ($2!$) according to the two digits that get dropped. In either interpretation there is no repetition of digits.
In comments below the OP, it's pointed out that you are allowing your eight-digit numbers to begin with a $0$, which may or may not be what the problem poser intended. As it turns out, however, it doesn't matter for the final result. If starting $0$'s are allowed, the probability is
$$10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\over10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10$$
whereas if starting $0$'s are not allowed, the probability is
$$9\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\over9\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10$$
In either case, cancellation leaves
$$9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\over10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10$$
Best Answer
Imagine a sequence of 16 random digits, uniformly distributed and chosen independently. What's the probability that there is no $7$ in it? The 1st digit is a non-7, and the 2nd digit is a non-7, and the 3rd digit is a non-7, etc. It's $\left(\dfrac{9}{10}\right)^{16}$. Similarly that's the probability that there is no $6$, etc.
One way to answer the whole question would be to apply the inclusion–exclusion principle:
There's no 0 or there's no 1 or there's no 2 or $\ldots\ldots$.
This is an inclusive "or".
\begin{align} & {}\qquad \Pr(\text{no $0$s or not $1$s or no $2$s or }\cdots) \\[10pt] & = \underbrace{\Pr(\text{no $0$s}) + \Pr(\text{no $1$s}) + \Pr(\text{no $2$s}) + \cdots}_\text{10 terms} \\[8pt] & \underbrace{{}- \Pr(\text{no $0$s and no $1$s}) - \Pr(\text{no $0$s and no $2$s}) - \cdots\cdots}_\text{45 terms} \\[8pt] & \underbrace{{}+ \Pr(\text{no $0$s and no $1$s and no $2$s}) + \Pr(\text{no $0$s and no $1$s and no $3$s})+ \cdots\cdots}_\text{120 terms} \\[8pt] & {} - \text{ etc., etc. } \cdots\cdots \\[12pt] & = 10\left(\frac{9}{10}\right)^{16} - 45\left(\frac{8}{10}\right)^{16} + 120\left(\frac{7}{10}\right)^{16} - \cdots\cdots \end{align}