All the numbers in the given expression are odd: $1 \times 3 \times 5 \times 7 \times 9 \times \dots \times 999$

Among these, every group of five consecutive odd numbers (e.g., 1–9, 11–19, 21–29, …, 991–999) has a product ending with 5. For instance, $1 \times 3 \times 5 \times 7 \times 9 = 945$ → unit digit = 5.

Similarly, $11 \times 13 \times 15 \times 17 \times 19$ also ends with 5, and so does each successive group. Thus, the overall multiplication becomes $5 \times 5 \times 5 \times \dots \times 5$, which always ends with 5.

Alternatively, multiplying any odd number by 5 always gives a number ending in 5. Since all factors here are odd, the final unit digit of the product must be $5$.