Question: let $k$ be a positive integer, $p$ a prime number, such that $p=3k+1$, $r<p$ be a positive integer, such that $2^{k+1}\equiv r\pmod p$, and $r\not\equiv 4,5\pmod 6$. Show that $$p\nmid k!+1.$$
It is well known $$\left(\dfrac{p-1}{2}\right)!^2+1\equiv 0\pmod p$$ where $p\equiv 1\pmod 4$, for my problem it seems hard to prove it.
It is known that each $p=3k+1$ has a unique representation of the form $$ p=\frac{L^2+27M^2}4, $$ up to sign change; see Wikipedia:.
The same Wikipedia page cites Gauss's theorem:
Under the same notation, we have $L\cdot k!^3\equiv 1\pmod p$.
[EDITED AND EXTENDED] As Geoff Robinson points out, this holds for an undetermined sign choice of $L$, so we need to work more; here it goes.
Assume now that $k!\equiv -1\pmod p$; this yields that $L\equiv -1\pmod p$, so either $L=-1$ or $|L|\geq p-1$. The latter cannot hold, as $4p<(p-1)^2+27$; so $L=-1$.
Thus, $4p=1+27M^2$. Writing $M=2m-1$, we arrive at the expansion $$ p=\frac{1+3M\sqrt{-3}}2\cdot \frac{1-3M\sqrt{-3}}2 =\bigl((3m-1)+3\omega(2m-1)\bigr)\bigl((3m-1)+3\bar\omega(2m-1)\bigr) $$ in the Eisenstein ring $\mathbb Z[\omega]$.
The number $$ \pi=(3m-1)+3\omega(2m-1) $$ is a primary Eisenstein integer, hence by the cubic reciprocity law (see the same Wiki page) we have $$ \left(\frac2\pi\right)_3=\left(\frac \pi2\right)_3 =\left(\frac{m+1+\omega}2\right)_3=:\alpha, $$ We have $\alpha=\omega$ if $m$ is odd, and $\alpha=\bar\omega$ if $m$ is even.
If $m$ is odd, we have $$ r\equiv 2^k\equiv \omega \equiv\omega-(2\omega+1)\pi =\omega-(\omega-(9m-5))=9m-5\pmod \pi. $$ Similarly, $r\equiv 9m-5\pmod{\bar\pi}$, and therefore $r\equiv 9m-5\pmod p$. Hence, $r=9m-5$ (notice that $9m-5<p$), and $r\equiv 4\pmod 6$. This is excluded by the conditions of the question.
If $m$ is even, we similarly get $$ r\equiv \bar\omega \equiv\bar\omega+(2\omega+1)\pi =-1-\omega+(\omega-(9m-5)) =4-9m\pmod\pi $$ and hence $r\equiv 4-9m\pmod p$, so $r=p+4-9m\equiv 5\pmod 6$. This is also excluded.
P.S. Pitifully, I have no access to that book, but Wikipedia claims that Gauss's theorem is Exercise 7.9 in Lemmermeyer's book:
Lemmermeyer, Franz (2000), Reciprocity Laws: from Euler to Eisenstein, Berlin: Springer, ISBN 3-540-66957-4
