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
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