Let's consider a factorial number $\frac{41!}{100}$. How can we find how many digits are zero starting from the last digit of the number?
If we multiply the factorial by $100$, then the number of $0$ digits increases by $2$ and we can work with $41!$ directly. We're looking for the largest $k$ such that
$$41!\equiv 0\pmod{10^k}$$
Furthermore, since $41!$ contains $2$ and $5$, a trailling zero exists. But I am not certain as to what can be done. Could you provide a feedback?
By Legendre's formula the number of trailing zeros of $41!$ is $$\nu_5(41!) = \left\lfloor \frac{41}{5} \right\rfloor + \left\lfloor \frac{41}{25} \right\rfloor = 9,$$ and dividing $41!$ by $100$ takes off two of those zeros, so there are $7$ trailing zeros.