Zeroes at the end of a factorial Write a program or function that finds the number of zeroes at the end of n! in base 10, where n is an input number (in any desired format).
It can be assumed that n is a positive integer, meaning that n! is also an integer. There are no zeroes after a decimal point in n!. Also, it can be assumed that your programming language can handle the value of n and n!.

Test cases
1
==&gt; 0

5
==&gt; 1

100
==&gt; 24

666
==&gt; 165

2016
==&gt; 502

1234567891011121314151617181920
==&gt; 308641972752780328537904295461

This is code golf. Standard rules apply. The shortest code in bytes wins.
Python 2, 27 bytes

f=lambda n:n and n/5+f(n/5)

The ending zeroes are limited by factors of 5. The number of multiples of 5 that are at most n is n/5 (with floor division), but this doesn't count the repeated factors in multiples of 25, 125, .... To get those, divide n by 5 and recurse.