1. The problem asks why $10! = 10 \times 9 \times 8!$. 2. The factorial of a number $n$, written as $n!$, is the product of all positive integers from $n$ down to 1. 3. By definition, $10! = 10 \times 9 \times 8 \times 7 \times \cdots \times 1$. 4. Notice that $8! = 8 \times 7 \times \cdots \times 1$. 5. Therefore, $10!$ can be rewritten as $10 \times 9 \times (8 \times 7 \times \cdots \times 1) = 10 \times 9 \times 8!$. 6. This shows the factorial property: $n! = n \times (n-1)!$. 7. So, $10! = 10 \times 9!$, and since $9! = 9 \times 8!$, it follows that $10! = 10 \times 9 \times 8!$. 8. This is a direct consequence of the factorial definition and the recursive nature of factorials.