Wilson's theorem asserts the following statement: $$(n-1)!\equiv -1\pmod n\Leftrightarrow n\text{ is prime.}\tag1$$ This means that $$\begin{align} n&{ \ \mid} \ \ (n-1)!+1 \\ \Leftrightarrow n&{ \ \mid} \ \ (n-1)!+1-n \\ &=(n-1)!-(n-1) \\ &= (n-1)\big((n-2)!-1\big) \\ \therefore n&{ \ \mid} \ \ (n-2)!-1 \\ \Leftrightarrow (n-2)!&\equiv 1\pmod n.\tag2\end{align}$$ Or, in the congruence, I could add $n$ to the right hand side, and then divide both sides by $n-1$, yeilding the same result.
I then asked myself, why wouldn't Wilson's theorem assert that $$(n-2)!\equiv 1\pmod n\Leftrightarrow n\text{ is prime,}$$ as opposed to $(1)$? It is more useful because $(n-2)! < (n-1)!$, so we can test this theorem with more primes without the factorial becoming too large as fast in $(1)$, and it is still just as interesting. However, I went here and found that it really is not that necessary to simplify the theorem.
Now, I have been trying to find some similar result for $(n-3)!$, and I found that $$(n-3)!\equiv \frac{n-1}{2}\pmod n\Leftrightarrow n\text{ is prime $> 2$.}$$
Main Question: If for some $k\in\mathbb{Z^+}$ and remainder $l_k\in\mathbb{Z}$, $$(n-k)!\equiv l_k\pmod n,\tag{$n$ is prime}$$ is there a pattern in the sequence $l_1, l_2, l_3,\ldots$?
Do there exist similar congruences for $(n-k)!$ such that $k > 3$?
Thank you in advance.
This post was inspired by this post.
Note that since $n$ is prime, we can define the following: $$y \equiv \frac{a}{x} \pmod{m} \iff n \mid xy-a \space (x,y,a\neq 0) $$
We can use the same properties of congruence such as adding, subtracting, multiplying and dividing by relatively prime constants.
Now, to find $(n-2)! \pmod{n} \space$, we basically need to find $\frac{1}{n-1} \pmod{n}$. We can instead write $n-1 \pmod{n}$ as $-1 \pmod{n}$. Thus, we have $\frac{1}{-1} \equiv -1 \pmod{n} $ which gives you that $(n-2)! \equiv 1 \pmod{n}$ using Wilson's Theorem.
The same idea goes on for $(n-k)!$. We are to evaluate $\frac{1}{-k} \pmod{n}$. For $k=1$, we trivially get $-1$. For $k=2$, considering $n$ as an odd prime, we know that $2 \mid n+1$. Thus, $\frac{n+1}{2}$ is an integer from which we can derive $(n-3)! \pmod{n}$.
Unfortunately, this process can no longer continue. A prime can be any value modulo another odd prime less than it except $0$. Thus, we cannot detect any pattern.