The number of zeroes at the end of 1000! has to be determined.

1000! = 1000*999*998*...*3*2*1

Each multiple of 5 in the factorial multiplied by an even number contributes one 0 at the end of the number equal to 1000!

There are 1000/5 = 200 multiples of 5.

The number of zeroes due to these is 200

In addition. each multiple of 25 multiplied by a multiple of 4 contributes an extra zero at the end of 1000!.

There are 1000/25 = 40 multiples of 25

The number of zeroes due to these is 40

Finally each multiple of 125 multiplied by a multiple of 8 contributes an extra zero at the end of 1000!

There are 1000/125 = 8 multiples of 125

The number of zeroes due to these is 8

The total number of zeroes in 1000! is the sum of zeroes due to the multiples of 5, the zeroes due to the multiples of 25 and the zeroes due to the multiples of 125. 200 + 40 + 8 = 248

**The number of zeroes at the end of 1000! is 248**