Question . Find all positive integer solutions to the equation below , $$(n-1)!+1=n^m$$ (i)observe that $n>1$ and $n$ is a prime number (if not we can choose a prime number $p<n$ such that $p|n$ this means that $p|(n^m-(n-1)!)=1$ which contradicts)
(ii)working on some primary primes gives us solutions $(n,m)=(2,1),(3,1),(5,2)$
(iii)I used some modular arithmetic modulo $p$ (Wilson's theorem)and $p+1$ and all that i got was parity of $m$ (being even) for $n>3$ ...
Can anybody help me prove this ...
I have poof if $p>5$ prime number,and $$(p-1)!+1=p^m$$ have no solution
Pf: since $p>5$,so $2<\dfrac{p-1}{2}<p-1$,so $$(p-1)^2=2\cdot\dfrac{p-1}{2}(p-1)|(p-1)!$$ if for $p>5$and postive integer $m$ such $$(p-1)!+1=p^m$$ then we have $$(p-1)^2|p^m-1\Longrightarrow p-1|p^{m-1}+p^{m-2}+\cdots+p+1$$ since $$p^{m-1}+p^{m-2}+\cdots+p+1\equiv m\pmod{p-1}$$ so $$p-1|m\Longrightarrow m\ge p-1$$
so$$p^m\ge p^{p-1}>(p-1)^{p-1}>(p-1)!$$ then $$p^m>(p-1)!+1$$ That's impossible.