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## Number of zeros at the end of n!

n! is defined as product of 1st n natural numbers. n! = n*(n-1)*(n-2)*….3*2*1, n must be natural (whole) number. Eg. 4!= 4*3*2*1 = 24 5!= 5*4*3*2*1 = 120 and so on. Note : 0! = 1 (explanation lies in theory of permutation & combination)