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Wednesday, July 21, 2021

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 

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