Find last 5 significant digits of 2017! Since there are less powers of $5$ than of $2$ and since $10 = 2 \cdot 5$, I counted the number of zeros in $2017!$: 
$\left \lfloor{ \frac{2017}{5^1}}\right \rfloor + \left \lfloor{ \frac{2017}{5^2}}\right \rfloor +\left \lfloor{ \frac{2017}{5^3}}\right \rfloor +\left \lfloor{ \frac{2017}{5^4}}\right \rfloor = 502$
Now I need to find $\frac{2017!}{10^{502}} \pmod{10^5}$
Using Chinese remainder theorem I split the $\pmod{10^5}$ into $\pmod{2^5}$ and $\pmod{5^5}$. Since there are a lot more powers of $2$ left, $\frac{2017!}{10^{502}}\equiv 0 \pmod{2^5}$
I am having trouble with getting $\frac{2017!}{10^{502}} \pmod{5^5}$. What would be the best way to compute it?
Clarification: significant digits - the ones that follow up before the zeros at the right side of the number.

I know how to do the last 3 significant digits, if that helps.
$4! = 24\\
\frac {9!}{5!} = 3024 \equiv 24\pmod {1000}$
$\frac {5n!}{5^n n!} = 24^n \pmod {1000}$ 
$2017! = \frac {2017!}{2015!}\frac {2015!}{5^{403}403!}\frac {5^{403}403!}{5^{80}80!}\frac {5^{80}80!}{5^{16}16!}\frac {5^{16}16!}{5^{3}3!} 3!\\$
$\frac {2017!}{10^502} \equiv 12^{502}\frac {2017!}{2015!}\frac {403!}{400!}\frac {16!}{15!}3!\pmod{1000}$
$144\cdot17\cdot16\cdot403\cdot402\cdot401\cdot16\cdot6 = 968\pmod{1000}$
