Johan Gunardi wrote: Find integer $ n$ with $ 8001 < n < 8200$ such that $ 2^n - 1$ divides $ 2^{k(n - 1)! + k^n} - 1$ for all integers $ k > n$. From condition we have $ n|k(n-1)!+k^n,\forall k>n$ . Choose $ k=n+1$ ,it is easy to check that $ n$ must be a prime and from Fermat's and Wilson's theorem , it is clear that all prime numbers satisfy . Now only need to find out all prime between $ (8001,8200)$ .

may ... $ 2^n - 1$ divides $ 2^{k(n - 1)! + k^n} - 1$ iff $ n|k(n - 1)! + k^n$ ?

jgnr wrote: Find integer $ n$ with $ 8001 < n < 8200$ such that $ 2^n - 1$ divides $ 2^{k(n - 1)! + k^n} - 1$ for all integers $ k > n$. We need $n|k(n - 1)! + k^n$ or $n|k^n$ for composites and $n|0$ for primes(which is clearly true) So if n was composite choose any $k \equiv -1 \pmod n$ and it won't work,so for this condition to hold we need n to be prime so the answer is $\boxed{21}$