##(p^k -1) \equiv X \mbox{(mod p)}## via Wilson's theorem

[DaTario]

TL;DR

Hi, is there a way to obtain $(p^k-1)! \equiv X \mbox{ (mod p)}$ for $X$ using Wilson's theorem: $[ (p-1)! \equiv -1 \mbox{(mod p)} ]$?

Hi All, being $p$ a prime number, is there a way to solve the congruence $(p^k-1)! \equiv X \mbox{ (mod p)}$ for $X$ using Wilson's theorem: $$ (p-1)! \equiv -1 \mbox{(mod p)} $$?

 

[fresh_42]

$k=1$ is Wilson and $k>1$ means $(p^k-1)!\equiv 0 \mod p$.

 

[DaTario]

Thanks!