Hi everybody,
Is there any theorem that shows:
$\displaystyle n^s-1\equiv 0 (mod (s+1))$
when s+1 is prime?
(this is valid for every n>1)
or need to be prooven?
This (in the form you have written) is not valid for every integer $\displaystyle n>1,$ only when $\displaystyle n$ and $\displaystyle s+1$ are coprime (in which case it is also valid for $\displaystyle n=1).$ However, if you re-write it as $\displaystyle n^{s+1}-n\equiv0\,(\bmod\,s+1)$ then it is valid for every integer $\displaystyle n$ (including 1).
It is more usual to write the prime $\displaystyle s+1$ as $\displaystyle p.$ Thus Fermat’s little theorem is usually presented as $\displaystyle n^{p-1}\equiv1\,(\bmod\,p)$ (which is valid if $\displaystyle \gcd(n,p)=1)$ or $\displaystyle n^p\equiv n\,(\bmod\,p)$ (which is valid for any integer $\displaystyle n).$