Since we know that the zeros at the end of any product are due to the presence of 10 as the factor of the product and the number of zeros depends upon the number of times 10 is involved. For example if there are seven 10's (i. e., 7 combinations of 5 × 2) in the n! then the number of zeros at the end of n will be 7.
But since, we have solved the same problem previous to this problem so we can conclude that the number of 10's in 1000! is 249. Hence the number of zeros at the end of the 1000! is 249.