2. (5 points) The product of $1 \times 2 \times 3 \times 4 \times \cdots \times 2014$ ends with $\qquad$ zeros.
【Solution】Solution: To calculate how many consecutive 0s are at the end of the product, we need to count the number of prime factor 5s contained in these 2014 numbers.

We can think about it this way: from 1 to 10, 5 contains one prime factor 5, and 10 contains one prime factor 5, meaning that the product of $1 \times 2 \times 3 \times 4$ $\times \cdots \times 10$ has a total of 2 zeros at the end;
5 contains 1 prime factor 5, 25 contains 2 prime factors 5, 125 contains 3 prime factors 5, and 625 contains 4 prime factors 5,

Calculating $2014 \div 5=402 \cdots 4, 2014 \div 25=80 \cdots 14, 2014 \div 125=16 \cdots 14, 2014 \div 625=3 \cdots 139$, $\therefore 1 \times 2 \times 3 \times 4 \times \cdots \times 2014$ has a total of $402+80+16+3=501$ consecutive 0s at the end. Therefore, the answer is 501.