We will use the "floor" function indicated by  ⌊x⌋ to denote the largest
integer that does not exceed x.  Sometimes this is indicated by int(x).

A factorial is a product of integers.  The only thing that will cause a 
zero at the end of a product of integers is a factor of 10, which is a 
pair of factors 2·5.

n! = 1·2·3·4·5·6···n

There are a lot more factors of 2 contained in the members of the sequence 

1,2,3,4,5,6,...,n 

than there are multiples of 5 contained in them.  So there will always be enough 
factors of 2 in n! to pair up with every factor of 5 in n! to cause the pair to
make a factor of 10 causing another zero at the end of 5! 

So we only need to count the total number of 5 factors contained in all the members 
of the sequence

1,2,3,4,5,6,...,n 

Each multiple of 51 contributes a 1st 5 factor.

There are n/⌊⌋ of those. Of those,

each multiple of 52 contributes a 2nd 5 factor.  

There are n/ ⌊⌋ of those. Of those,

each multiple of 53 contributes a 3rd 5 factor.

There are n/ ⌊⌋ of those. Of those,
...

n/ ⌊⌋ will become 0 when 5k
exceeds n.  

So the number of 0's at the end of n! is

⌊⌋ + ⌊⌋ + ⌊⌋ + ... + ⌊⌋, 

where M is the largest power of 5 that does not exceed n, or even
larger since values of k > M will all be 0.

We can calculate M, which is the largest value of k such that

5k ≦ n

ln(5k) ≦ n

k·ln(5) ≦ n

      k ≦ 

So M = ⌊⌋

You can write the result as a summation:

Number of 0's at end of n! = ⌊⌋, where M = ⌊⌋ 

Edwin