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
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