Does $625!$ have $156$ zeros at the end? Someone wrote that $625!$'s last $156$ digits are zeros because $125+25+5+1=156$.
If it's true that $625!$ has $156$ zeros at the end, how does "$125+25+5+1=156$" prove it?
 A: When you factorize $625!$, you are interested in the number of $10$s you can divide it by until it is no longer possible to divide evenly. 
But $10=2\cdot 5$, so you actually want to know how many $2$s and $5$s are in the prime factorization of $625!$. You will quickly realize if you start listing the product that there are many many more factors of $2$s than there are $5$s, which means that the number of zeroes at the end is exactly equal to the number of $5$s in the prime factorization of $625!$.
How many $5$s are there? Well every multiple of $5$ in the product will count as one five. But every multiple of $25$ will count as an extra $5$, and every multiple of $125$ will count an extra on top of that, and every multiple of $5^4 = 625$ will count a fourth time.
There are $125$ positive multiples of $5$ less than or equal to $625$, $25$ multiples of $25$ less than or equal to $625$, $5$ multiples of $125$ less than or equal to $625$, and just the one multiple of $625$.
This gives your 
$$125 + 25 + 5 + 1 = \boxed{156}$$
zeroes.
