Getting ready for 2015

Note, from Wilson's theorem, that $2016! \equiv -1 \pmod {2017}$ . Additionally, if $2015! \equiv x \pmod {2017}$ , we have $2016! \equiv -x \pmod {2017}$ , so $x \equiv 1$ .

Hence, $\begin{aligned} 2015!^{2015!} &\equiv 1^{2015!} \\&\equiv \boxed{1} \pmod {2017}. \end{aligned}$

Observe that $2016 | 2015!$ and that $2017$ is a prime number.

By Fermat's little theorem, $a^{p-1} \equiv 1 \mod p$ when $\gcd(a, p) = 1$ .

Setting $a = 2015!$ and $p = 2017$ , we have:

$2015!^{2015!} = (2015!^{2016})^{2015!/2016} \equiv 1^{2015!/2016} \equiv \boxed{1} \mod 2017$ .

Note that 2017 is prime, so $\phi(2017)=2016$ . Therefore, we find $2015!\mod 2016$ , and as 2016 is $2*1008$ , $2015!\equiv0\mod2016$ . Therefore, the modulo is equivalent to $2015!^0\mod 2017=1\mod2017$ .