2021/053) Prove that $n^{n-1}-1$ is divisible by $(n-1)^2$ for integer n

Replacing n by k+ 1 we need to prove

$(k+1)^k-1$ is divisible by $k^2$

If k is less than 2 it is obvious so let us check for k greater than 2

We have $(k+1)^k - 1 =\sum_{n=0}^k {k \choose n} k^n - 1$

$= 1 + k . k + \sum_{n=2}^k {k \choose n} k^n -1 $

$ =  k . k + \sum_{n=2}^k {k \choose n} k^n$

Each term is divisible by $k^2$ and hence the expression