How many zeroes?

The number of trailing zeros $z_n$ of $n!$ for $n>5$ is given by:

$\quad z_n = \lfloor \frac {n}{5} \rfloor + \lfloor \frac {n}{5^2} \rfloor + \lfloor \frac {n}{5^3} \rfloor +...$

where $\lfloor \dot{} \rfloor$ is the greatest integer function.

Therefore the number of trailing zeros of $100!$ is:

$\quad z_{100} = \lfloor \frac {100}{5} \rfloor + \lfloor \frac {100}{25} \rfloor = 20 + 4 = \boxed {24}$