2021/104) Prove that for positive integer n we have $n^2 | (n+1)^n-1$

We have

$(n+1)^n - 1$

$= \sum_{k=0}^{n}  {n \choose k} n^{n-k} -1 $

$= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} + {n \choose n-1} n^{n-(n-1)} + {n \choose n} n^{n-n} -1$

$= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} +n * n + 1 - 1$

$= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} +n^2$

now each term in sum is having $n^2$ as a term and hence the expression is divisible by $n^2$