The number is so cute

As the number of trailing zeros of $n!$ is given by $z_n = \displaystyle \sum_{k=1}^{\left \lfloor \log_5 n \right \rfloor} \left \lfloor \frac n{5^k} \right \rfloor$ , the number of trailing zeros of a product of factorials is the sum of $z_n$ . Therefore, we have:

$\begin{aligned} Z_N & = \sum_{n=1}^{2017} z_n \\ & = \sum_{n=1}^{2017} \sum_{k=1}^{\left \lfloor \log_5 n \right \rfloor} \left \lfloor \frac n{5^k} \right \rfloor \\ & = \sum_{n=1}^{2017} \left(\left \lfloor \frac n5 \right \rfloor + \left \lfloor \frac n{25} \right \rfloor + \left \lfloor \frac n{125} \right \rfloor + \left \lfloor \frac n{625} \right \rfloor \right) \end{aligned}$

Evaluate the summations separately:

$\begin{aligned} S_1 & = \sum_{n=1}^{2017} \left \lfloor \frac n5 \right \rfloor \\ & = \left \lfloor \frac 15 \right \rfloor + \left \lfloor \frac 25 \right \rfloor + \left \lfloor \frac 35 \right \rfloor + \left \lfloor \frac 45 \right \rfloor + \left \lfloor \frac 55 \right \rfloor + \left \lfloor \frac 65 \right \rfloor + \left \lfloor \frac 75 \right \rfloor + \left \lfloor \frac 85 \right \rfloor + \left \lfloor \frac 95 \right \rfloor \\ & \quad + \left \lfloor \frac {10}5 \right \rfloor + \left \lfloor \frac {11}5 \right \rfloor + \left \lfloor \frac {12}5 \right \rfloor + \left \lfloor \frac {13}5 \right \rfloor + \left \lfloor \frac {14}5 \right \rfloor + \left \lfloor \frac {15}5 \right \rfloor + \left \lfloor \frac {16}5 \right \rfloor + \cdots \\ & \quad \cdots + \left \lfloor \frac {2012}5 \right \rfloor + \left \lfloor \frac {2013}5 \right \rfloor + \left \lfloor \frac {2014}5 \right \rfloor + \left \lfloor \frac {2015}5 \right \rfloor + \left \lfloor \frac {2016}5 \right \rfloor + \left \lfloor \frac {2017}5 \right \rfloor \\ & = 0+0+0+0+1+1+1+1+1 +2+2+2+2+2+3+3+\cdots+402+402+402+403+403+403 \\ & = \sum_{n=1}^{\left \lfloor \frac {2017}5 \right \rfloor -1} 5n +\left(2018 - 5\left \lfloor \frac {2017}5 \right \rfloor \right) \left \lfloor \frac {2017}5 \right \rfloor \\ & = \sum_{n=1}^{402} 5n + \left(2018 - 2015 \right)403 \\ & = \frac {5(402)(403}2 + 3(403) \\ & = 406224 \end{aligned}$

Similarly,

$\begin{aligned} S_2 & = \sum_{n=1}^{2017} \left \lfloor \frac n{25} \right \rfloor = \frac {25(79)(80)}2 + (2018-2000)80 = 80440 \\ S_3 & = \sum_{n=1}^{2017} \left \lfloor \frac n{125} \right \rfloor = \frac {125(15)(16)}2 + (2018-2000)16 = 15288 \\ S_4 & = \sum_{n=1}^{2017} \left \lfloor \frac n{625} \right \rfloor = \frac {625(2)(3)}2 + (2018-1875)3 = 2304 \end{aligned}$

$\implies Z_N = S_1 +S_2 + S_3+S_4 = 406224 + 80440 + 15288 + 2304 = \boxed{504256}$