Concatenating Factorials with Astounding Powers

Euler's theorem says ${a}^{\phi(n)} = 1 mod n$ where a & n are co-prime.

Therefore let's find out the totient function of n or 4056195 or 2013 2015 = 3 11 61 5 13 31

$\phi(p) = p-1$ and $\phi(a*b) = \phi(a)*\phi(b)$

hence,

\(\phi(4056195) = 2*10*60*4*12*30

Now, \(2013!*2015!\) has to have all these numbers in it.

Therefore, ${2}^{2013!*2015!} = 1 mod 4056195$

Z = 1, so, ${z}^{2014}(mod 1000)$ = 1 mod 1000 = 1

just factorzie 4056195 and use ferrmet theorm