Advanced happy new year!

Using Trailing Number of Zeros , we want

$2016=\left \lfloor \frac{n}{5} \right \rfloor+ \left \lfloor \frac{n}{5^2} \right \rfloor+ \left \lfloor \frac{n} {5^3} \right \rfloor+\ldots$

Using an approximation from Geometric Progression Sum ,

$2016=\frac{\frac{n}{5}}{1-\frac{1} {5}}$

$2016=\frac{n}{4}\Rightarrow n=8064$

Number of zeroes in $8064!=2012$ .

Therefore to get four more zeroes (i.e. $2016$ zeroes) the answer must be $\boxed{8075}$ .

Quicker Approximation

$x=4n$ where $x!$ has $n$ zeroes.

Implementation Trailing Number of Zeros , I use C++

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#include<bits/stdc++.h>
using namespace std;

int main(){
long long a,b,c;

a=5;
cin>>b;  //INPUT=8075
while(a<b){
    c=c+(long long)ceil(b/a);
    a*=5;
}

cout<<c<<endl; //OUTPUT=2016

}

nice and brief solution

No. of trailing zeroes for any number say $n$ is $[\frac{n}{5}]+[\frac{n}{25}]+[\frac{n}{125}]+...$ where $[.]$ is greatest integer function for $n€N$ using this I got minimum value of n equal to $\boxed{8075}$