Ta có: $n ≡ 1$ (mod n - 1)
=> $n^x ≡ 1$ (mod n - 1)
=> $n^{n-1} + n^{n-2} + n^{n-3} + ... + n^3 + n^2 + n$
$≡ 1 + 1 + 1 + ..... + 1$
$= n - 1$
$≡ 0$ (mod n - 1)
=> $n^{n-1} + n^{n-2} + n^{n-3} + ... + n^3 + n^2 + n$ chia hết cho n - 1.
Bạn tham khảo nhé.