8. Suppose a random number selector can only choose a number from $1,2, \cdots, 9$ each time, and it makes the selection with equal probability. Then, after $n(n>1)$ selections, the probability that the product of the $n$ selected numbers is divisible by 10 is $\qquad$
8. $1-\frac{8^{n}+5^{n}-4^{n}}{9^{n}}$.

To make the product of $n$ numbers divisible by 10, there must be one number that is 5 and one number that is even. The total number of ways to choose $n$ times is $9^{n}$, among which:
(1) The number of ways to choose without taking 5 each time is $8^{n}$;
(2) The number of ways to choose without taking an even number each time is $5^{n}$;
(3) The number of ways to choose from $\{1,3,7,9\}$ each time is $4^{n}$.

Clearly, the methods in (3) are included in both (1) and (2), so the probability that the product of $n$ numbers is divisible by 10 is $1-\frac{8^{n}+5^{n}-4^{n}}{9^{n}}$.