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Divisibility of \(n^{m} – n\)

It is quite easy to show that \(n^{2} – n\) always has 2 as a factor (i.e. can be divided by two).

It is fairly easy to show that \(n^{3} – n\) always has 3 as a factor, and that it also has 2 as a factor.

  • Is it true that \(n^{m} – n\) always has m as a factor?
  • Is it true that \(n^{m} – n\) always has m! as a factor?

Both n and m are positive integers.

Note by Johan Falk
2 years, 11 months ago

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