In a recent article, Emmanuel Lecouturier proves a generalization of the following surprising result: for a Mersenne prime $N = 2^p - 1 \ge 31$, the element $$ S = \prod_{k=1}^{\frac{N-1}2} k^k $$ is a $p$-th power modulo $N$, and observed that he did not know an elementary proof. Neither do I.

Numerical experiments suggest that $s_p$ is actually a $6p$-th power modulo $N$. I can't even see why it is a quadratic residue, i.e., why the following result (not proved in the article cited) should hold: $$ T = \prod_{k=1}^{\frac{N+1}4} (2k-1) $$ is a square mod N.

For arbitrary primes $N \equiv 3 \bmod 4$, the following seems to hold: $$ \Big(\frac{T}{N} \Big) = \begin{cases} - (-1)^{(h-1)/2} & \text{ if } N \equiv 3 \bmod 16, \\ - 1 & \text{ if } N \equiv 7 \bmod 16, \\ (-1)^{(h-1)/2} & \text{ if } N \equiv 11 \bmod 16, \\ + 1 & \text{ if } N \equiv 15 \bmod 16, \end{cases} $$ where $h$ is the class number of the complex quadratic number field with discriminant $-N$. This suggests a possible proof using L-functions (i.e. using methods in (Congruences for L-functions, Urbanowicz, K.S. Williams) and explains the difficulty of finding an elementary proof.

My questions:

- Is this congruence for $(T/N)$ known?
- How would I start attacking this result using known results on L-functions?