**I have a problem. I want to find factorial of big numbers.**

Ex: 1555! = ?.
195! = ?.

My main problem is that I want to know the exact number of ending 0's of the factorial numbers.

I use the following formula:
 (m!)^n = m! = 2\*10^(n-1) + 2^2 \* 10^(n-2) + ------- + 2^n.

with this I can solve the other factorials for number of ending 0's like this.

100!= 2\*10^1 + 2^2\*10^0 = 20+4 = 24

100! has 24 ending 0's as per this calculation.

But, then I got other problem, 

Ex: For 95!

i) 95! = (100 - 5)! = 24 - 2\*5^(1-1) = 24 - 2 = 22 => 95! has 22 0's.

ii) 95! = (90 + 5)! = 9\*(2\*10^0) + 2\*5^0)= 18+2 = 20 => 95! has 20 0's.

this is my problem. By using the above formula I got two different answers and I am confused, I don't get the perfect answer so please help me to find it.

***Thank you...***