Find $$(n,k)$$ ( n,k are positive integers ) pairs that satisfy the following equation: $$n^k-1=(n-1)!$$
I found this question in a discrete math test book in number theory section and tried solving it using Wilson's theorem:
$$(p-1)!+1$$ is divisible by n if and only if p is a prime number.
Using this theorem we can say that if n is a prime number then the remainder of $n^k-1$ divided by n equals the remainder of $(n-1)!$ divided by n but I don't know how to proceed from here.
