Find the remainder when $\frac{13!}{7!}$ is divided by $17$. How to find the remainder when $\frac{13!}{7!}$ is divided by $17$?
I started with Wilson's Theorem which states that every prime $p$ divides $(p-1)!+1$. That is $(p-1)!=-1 mod(p)$. 
$16!=-1 mod(17)$
Kindly help me how to get to $\frac{13!}{7!}$.
 A: $$\dfrac{13!}{7!}=\prod_{r=8}^{13}r\equiv(-1)^64\cdot5\cdot6\cdot7\cdot8\cdot9$$
$$\equiv3\cdot8\cdot4\equiv11\pmod{17}$$
A: By Wilson's theorem $16!\equiv -1\pmod{17}$ and since $17$ is a prime of the form $8k+1$ and the square roots of $-1$ in $\mathbb{F}_{17}^*$ are $4$ and $13$, $8!\equiv 13\pmod{17}$ and $7!\equiv 8\pmod{17}$. By using the notation $\frac{1}{a}$ for the inverse of $a$,
$$ \frac{13!}{7!}\equiv\frac{16!}{(-1)(-2)(-3)8}\equiv\frac{1}{48}\equiv\frac{1}{-3}\equiv -6\equiv\color{red}{11}\pmod{17}.$$
A: $\frac{13!}{7!}=8\cdot9\cdot10\cdot11\cdot12\cdot13\equiv (-9)\cdot(-8)\cdot(-7)\cdot(-6)\cdot(-5)\cdot(-4) \mod(17) \equiv 36 \cdot 56\cdot30 \mod(17) \equiv 2\cdot5\cdot13\mod(17) \equiv 130 \mod(17) \equiv 11 \mod(17)$
Apologies for the badly-written solution
A: Touching on @ProfessorVector's hint
$$8 \times 9=72\equiv 4,$$
then
$$4\times 10=40\equiv 6,$$
then
$$6\times 11=66\equiv 15,$$
then
$$15\times 12=180\equiv 10,$$
then
$$10\times 13=130\equiv 11.$$
Thus, in mod $17$
$$\frac{13!}{7!}=11.$$
