If $N$ is a multiple of $100$, $N!$ ends with $\left(\frac{N}4-1 \right)$ zeroes. Did certain questions about factorials, and one of them got a reply very interesting that someone told me that it is possible to show that

If $N$ is a multiple of $100$, $N!$ ends with $\left(\frac{N}4-1 \right)$ zeroes.

Sought such a demonstration, not found when trying to do there, I thought of induction, but did not fit, I thought of several ways, but did not get any progress, would like to know how to do, or to see a demonstration of this.

I thought about trying to calculate the following series $$\sum_{k=0}^{\infty} \left[\frac{n}{5^k}\right]$$ only that I was in doubt because it is using the quotients $\left[\frac{n}{5^k}\right]$

 A: Hint:
A number ending with zero is divisible by $10=2\times5$. Hence we have to find the multiplicities of $2$ and $5$. However the multiplicity of $2$ clearly dominates that of $5$. Hence it suffices to find the multiplicity of $5$. 
Number of multiples of $5$ = $\left\lfloor\frac{N}{5}\right\rfloor$  
Number of multiples of $25$ = $\left\lfloor\frac{N}{5^2}\right\rfloor$
Number of multiples of $125$ = $\left\lfloor\frac{N}{5^3}\right\rfloor$  ...
A: The correct formula for the number of trailing zero digits in $n!$ is
$$
\frac{n-\sigma_5(n)}{4}
$$
where $\sigma_5(n)$ is the sum of the base-$5$ digits of $n$. So the formula given to you is only correct if the sum of the base-$5$ digits of $n$ is $4$. Since
$$
\begin{array}{l}
100_\text{ten}=400_\text{five}\\
200_\text{ten}=1300_\text{five}\\
300_\text{ten}=2200_\text{five}\\
400_\text{ten}=3100_\text{five}\\
500_\text{ten}=4000_\text{five}
\end{array}
$$
the formula given to you works for each of them. However, $600_\text{ten}=4400_\text{five}$, so we get that the number of trailing zero digits will be
$$
\frac{600}{4}-2
$$
one less than the formula given to you.
A: It's not true. Consider $N = 600$.
$$\left\lfloor\frac{600}{5} \right\rfloor + \left\lfloor \frac{600}{25}\right\rfloor + \left\lfloor \frac{600}{125}\right\rfloor = 120 + 24 + 4 = 148 = \frac{600}{4} - 2.$$
