2015!

Wilson told us,

$2016! \equiv -1 \pmod{2017}$

or, $2016! \equiv 2016 \pmod{2017}$

or, $2015! \equiv 1 \pmod{2017}$ [Dividing both sides by $2016$ ]

The remainder is $\boxed{1}$ ,when $2015!$ is divided by 2017.

We know from Wilson's theorem that

(p-2)! $\equiv$ 1 (mod p)

2017 is prime so 2015! will be congruent to 1 (mod 2017).

Directly follows from wilsons theorem and the fact that $2016=-1 \mod 2017$ .

Hence, $\boxed{1}$