Find $e \in \{0,1,\ldots,22\}$ such that the product $\prod_{i=6}^{18} i$ is congruent to $e$ modulo $23$ Find $e \in \{0,1,\ldots,22\}$ such that the product $\prod_{i=6}^{18} i$ is congruent to $e$ modulo $23$.
I have completed a question similar to this before so I thought I knew how to do this but I believe my answer is wrong.
In order to complete this question these are the steps I took:
$$12 \cong -11 \pmod{23}$$
$$13 \cong -10\pmod{23}$$
$$14 \cong -9\pmod{23}$$
$$15 \cong -8\pmod{23}$$
$$16 \cong -7\pmod{23}$$
$$17 \cong -6\pmod{23}$$
$$(6)(7)(8)(9)(10)(11)(-11)(-10)(-9)(-8)(-7)(-6)$$
$$(-)[6\cdot11][7\cdot10][8\cdot9][9\cdot8][10\cdot7][11\cdot6]\pmod{23}$$
$$(-)(66)(70)(72)(72)(70)(66)$$
$$66=20\pmod{23}$$
$$70=1\pmod{23}$$
$$72=3\pmod{23}$$
$$=(-)(20)(1)(3)(3)(1)(20)$$
The next parts are what confuses me because I then use: $[20\cdot3]^2(1)^2 = 60^2$ and $60\cong14\pmod{23}$ so $(-)(14)^2\pmod{23} =(-)(196)\pmod{23}=12$ and then I finally get that $-12\pmod{23}\cong 11\pmod{23}$ as my final answer. I was told that this is wrong and I am unsure why. I am fairly certain the errors begin after I get:${}=(-)(20)(1)(3)(3)(1)(20).$

Recall for a prime $p$, $(p-1)! \equiv_{p} -1 $. It follows
$$2 \cdot 3 \cdot 4 \cdot 5\cdot \left( \prod_{i=6}^{18}  i\right)\cdot (-4) \cdot (-3) \cdot (-2) \cdot (-1)\equiv_{23} (-1)$$
$$\Rightarrow 5 \cdot \left(\prod_{i=6}^{18} i \right) \equiv_{23} -1 \quad \Rightarrow \prod_{i=6}^{18} i \equiv_{23} 9 $$
