Zeroes everywhere

Since the factors 5 and 2 are needed to make a trailing zero, we can take the floor function of how many times the powers of 5 divide 60 (since they are less common than powers of 2).

$\left \lfloor \frac{60}{5} \right \rfloor + \left \lfloor \frac{60}{5^{2}} \right \rfloor = 12+2 =\boxed{14}$

In general, the number of trailing zeroes in x! = $\left \lfloor \frac{x}{5} \right \rfloor + \left \lfloor \frac{x}{5^{2}} \right \rfloor + \left \lfloor \frac{x}{5^{3}} \right \rfloor ... + \left \lfloor \frac{x}{5^{n}} \right \rfloor where \frac{x}{5^{n+1}} < 1 < \frac{x}{5^n}$

NB: $\left \lfloor x \right \rfloor$ is the floor function of x and is used to denote the largest integer not greater than x