Intermediate Mathematics-2.

$2016^{2016} \mod 2017 = ?$

$2016 \equiv -1 (\mod 2017)$

$(2016^{2016} \mod 2017) = ((-1)^{2016} \mod 2017) = (1 \mod 2017) = 1$

Since 2017 is a prime, $\gcd(2016, 2017) = 1$ and we can apply Euler's theorem . Since 2017 is a prime, Euler's totient function $\phi (2017) = 2016$ . Therefore, $2016^{2016} \equiv 2016^{2016 \bmod \phi(2017)} \equiv 2016^{2016 \bmod 2016} \equiv 2016^0 \equiv \boxed 1 \text{ (mod 2017)}$ .

$2017$ is prime. Fermat's Little Theorem states that $a^{p-1}\equiv 1\bmod p$ for any prime $p$ and all $a$ not divisible by $p$ .

$2016^{2016}=2016^{2017-1} \equiv \boxed{1}\bmod 2017$