$n^n - n^2 + n - 1$
$= n^n - 1 - n^2 + n$
$= (n-1)(n^{n-1} + n^{n-2} +\dots +n +1) - n(n-1)$
$= (n-1)(n^{n-1} + n^{n-2} +\dots +n +1 - n)$
$= (n-1)[n^{n-1} + n^{n-2} +\dots +n +1 + \underbrace{(-1-1-1-\dots-1-1)}_{\text{n số -1}}]$
$= (n-1)[(n^{n-1} - 1) + (n^{n-2}-1) +\dots + (n-1) + 1 - 1]$
$= (n-1)[(n^{n-1} - 1) + (n^{n-2}-1) +\dots + (n-1)]$
Ta có:
$a^n - b^n \quad\vdots \quad a - b\quad\forall n \in \Bbb N;\, a,b\in\Bbb Z$
Do đó:
$n^{n-1} -1 = n^{n-1} - 1^{n-1}\quad\vdots \quad n -1$
$n^{n-2} - 1 = n^{n-2} - 1^{n-2}\quad\vdots \quad n -1$
$\cdots$
$n-1\quad\vdots \quad n -1$
$\Rightarrow (n^{n-1} - 1) + (n^{n-2}-1) +\dots + (n-1) \quad \vdots \quad n -1$
$\Rightarrow (n-1)[(n^{n-1} - 1) + (n^{n-2}-1) +\dots + (n-1)] \quad \vdots \quad (n -1)^2$
Hay $n^n - n^2 + n -1 \quad \vdots \quad (n -1)^2$ (đpcm)
