Show $[(p-1)!]^{p^{n-1}} \equiv -1 $ (mod $p^n$) for n $\in \mathbb N$, by induction. p a prime and p>2.
I can't seem to prove the inductive step for this. Would appreciate help.
My approach has been:
n=1 is just from Wilson.
Assume true for n=m: $[(p-1)!]^{p^{m-1}} \equiv -1 $ (mod $p^m$)
Then,
$[(p-1)!]^{p^{(m+1)-1}} = ([(p-1)!]^{p^{m-1}})^p \equiv (-1)^p \equiv -1$ (mod $p^m$)
But, how do I get this to say anything in terms of mod $p^{m+1}$? Since I need the RHS to end up as: -1 (mod $p^{m+1}$).
One thing I could draw from this congruence is that $[(p-1)!]^{p^{(m+1)-1}}$ is not a multiple of p, since multiples of p must be greater than -1 apart from each other.
Hence, GCD($[(p-1)!]^{p^{(m+1)-1}}, p^{m+1})=1.$ I wasn't sure how I might use this.
Alternatively, I could express it as: $[(p-1)!]^{p^{(m+1)-1}} = [(p-1)!]^{p^m}$. But this didn't seem the right way to go about it, since by cancelling the -1+1, there doesn't seem to be any way to use the inductive hypothesis/assumption above.
Another useful result might be that: GCD((p-1)!,$p^{m+1}$)=1.