Find the remainder when \[9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}\] is divided by $ 1000$.
1. We need to find the remainder when the product \(9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}\) is divided by 1000. 

2. Notice that each number in the product can be written in the form \(10^k - 1\), where \(k\) is the number of digits in the number. For example:
   - \(9 = 10^1 - 1\)
   - \(99 = 10^2 - 1\)
   - \(999 = 10^3 - 1\)
   - and so on up to \(\underbrace{99\cdots9}_{\text{999 9's}} = 10^{999} - 1\).

3. We can use the property of modular arithmetic that states if \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then \(a \times c \equiv b \times d \pmod{m}\). 

4. We will compute the product modulo 1000. First, observe that for \(k \geq 3\), \(10^k \equiv 0 \pmod{1000}\). Therefore, for \(k \geq 3\), \(10^k - 1 \equiv -1 \pmod{1000}\).

5. The product can be broken down as follows:
   \[
   9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}} \equiv (10^1 - 1) \times (10^2 - 1) \times (-1)^{997} \pmod{1000}
   \]

6. Simplify the first two terms:
   \[
   9 \equiv 9 \pmod{1000}
   \]
   \[
   99 \equiv 99 \pmod{1000}
   \]

7. Combine the first two terms:
   \[
   9 \times 99 = 891 \pmod{1000}
   \]

8. Now, multiply by \((-1)^{997}\):
   \[
   891 \times (-1)^{997} \equiv 891 \times (-1) \equiv -891 \pmod{1000}
   \]

9. Convert \(-891\) to a positive remainder:
   \[
   -891 \equiv 1000 - 891 \equiv 109 \pmod{1000}
   \]

The final answer is \(\boxed{109}\).