I really have to get working on my report now, but I couldn't resist to write a small excel program to numerically work out the number of trailing zero's. Of course excel can't handle sufficiently large numbers to describe the number 100! accurately. But you can proof the number of trailing zero's by multiplying each neightbouring pair of numbers and factoring out any multiples of 10 before executing the next series of neighbouring multiplications.

You stop when all (remaining) neighbouring pairs both have a trailing digit out of the series 1,2,3,4,6,7,8 or 9 where the left over series can never again produce another trailing zero (another 10 that can be factored out). This happens after defactorizing the 2nd series.

Totalling the number of factored out 10's results in 24, ergo the number 100! has 24 trailing zero's.

See the table attached : (POS = number is now Part Of Set with trailing digit 1, 2, 3, 4, 6, 7, 8 or 9)