Find the smallest positive integer x such that 2015! ≡ x (mod 2017) Q. The next year that is a prime is 2017. Find the smallest positive integer x such that 
2015! ≡ x (mod 2017).
So, this is what I have;
By Wilson’s theorem, (2017-1)! ≡ -1 (mod 2017) ⇒ 2016! ≡ -1 (mod 2017) ⇒ 
2016∙2015! ≡ -1 (mod 2017) ⇒
not sure how to do the modular arithmetic to have just 2015! on the left side..
 A: As $2017$ is prime, using Wilson's Theorem $$2016!\equiv-1\pmod{2017}$$
$$\iff 2015!\cdot 2016\equiv-1\pmod{2017}$$
$$\iff 2015!\cdot (-1)\equiv-1\pmod{2017}\text{ as }2016\equiv-1\pmod{2017}$$
So, $\cdots$
A: Hint:
$2016$ is relatively prime to $2017$
If, $c$ is relatively prime to $p$, then,
$$ac\equiv bc\pmod{p}\iff a\equiv b\pmod{p}$$
Also, $$2016\equiv -1\pmod{2017}$$

Also, if I may lengthen this on purpose, then let's do this without Wilson's Theorem.

Consider $2 \le a\le 2015 $. To each such $a$, we associate it's unique inverse $\overline a$, i.e, $a\overline a\equiv1\pmod{2017}$. Note that $a\ne \overline a$, because then $a^2\equiv 1\pmod{p}$ which would mean $a\equiv \pm 1\pmod{p}$ which is not possible for the $a$ in consideration. Thus in multiplying all $a\in \{2,3,\cdots,2015\}$ we pair them with their inverses, and the net product is $1$. 
$$2015!\equiv 1\pmod{2017}$$
QED
A: Your solution is almost done. Notice that $2016 \equiv -1 \text{ }mod(2017)$ so $$2015!(-1)\equiv (-1) mod (2017)\implies 2015!\equiv 1$$
