Đáp án `:`

`n^n-n^2+n-1=n^n-n-(n^2-2n+1)=n^n-n-(n-1)^2`

Ta có `:`

`n^n-n=n(n^(n-1)-1)=n(n-1)(n^(n-2)+n^(n-3)+....+n+1)`

Lại có `:`

`n^(n-2)+n^(n-3)+...+n+1` $\equiv$ `1+1+....+1` (có `n-1` chữ số `1`) (mod `n-1`)

`->n^(n-2)+n^(n-3)+....+n+1` $\equiv$ `n-1` (mod `n-1`)

`->n^(n-2)+n^(n-3)+....+n+1=k(n-1)` `(k in NN**)`

`->n(n-1)(n^(n-2)+n^(n-3)+....+n+1)=kn (n-1)^2` `(k in NN**)`

`->n^n-n=kn (n-1)^2` `(k in NN**)`

`->n^n-n` $\vdots$ `(n-1)^2` 

Hay `n^n-n^2+n-1`  $\vdots$ `(n-1)^2` (đpcm)