8. 1500 products are packed into 15 boxes before leaving the factory. After the boxes are sealed, it is found that one product is still not packed into any box. Due to the high sealing requirements, it is not allowed to open each box for inspection. Someone suggests using a scale to weigh each box (since the box with one less product will be lighter). By the usual method, each box would need to be weighed once, requiring 14 weighings to identify the box with the missing product. Is there a way to reduce the number of weighings while still being able to identify the box with the missing product?
8. It is certain that the box with less content can be found with just 4 weighings. Number the 15 boxes of products as $1,2, \cdots, 15$, and represent $1 \sim 15$ in binary (omitted).

It is easy to know that the numbers from 1 to 15 can be represented in binary using 4 digits. By grouping the box numbers where each digit is "1", we can get 4 groups: $1,3,5,7,9,11,13,15 ; 2,3,6,7,10,11,14,15 ; 4,5,6,7,12,13, 14,15 ; 8,9,10,11,12,13,14,15$.

Weigh each group of boxes once. If the weight is normal, record it as 0; if the weight is lighter, record it as 1. After 4 weighings, a binary number $x=(\stackrel{\oplus}{\times} \times \stackrel{\otimes}{\times} \times \stackrel{\oplus}{\times})_{2}$ will be obtained, which is the box number of the box with less content. For example, if the results of the first and third groups are lighter, then $x=(0101)_{2}=5$, indicating that the lighter box (i.e., the box with less content) is definitely the 5th box.