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I am trying to solve the following question:

For each positive integer $n$, strictly larger than 1, it holds that $n^{n-1}-1$ is divisible by $(n-1)^2$.

This question appears in a chapter on the binomial theorem. I tried solving it by induction, with not much of a result. Then I decided to take another approach: just use Euclidean division. I got the following result: $$n^{n-1}-1 = (n-1)(n^{n-2} + n^{n-3} + \ldots + n +1)$$ hence I need to prove that the second factor is divisible by $n-1$. Here, I applied Euclidean division and I think the result is something like $$n^{n-2} + n^{n-3} + \ldots + n +1 = (n-1)(n^{n-3} + 2n^{n-4} + \ldots + )$$ I have trouble finding the correct terms.

Also, this way of solving does not use the binomial theorem. The exercises in this chapter are specifically on this theorem and on some binomial coefficient-identities.

My question is: can anyone explain me how I could solve this using the binomial theorem? (also, is what I have done so far correct and how to proceed?)

nonuser
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Student
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2 Answers2

6

Hint: Note that $$n\equiv 1 \pmod{n-1}$$

so

$$n^{n-2} + n^{n-3} + \ldots + n +1 \equiv \underbrace{1+1+...+1}_{n-1}\equiv 0 \pmod{n-1}$$

nonuser
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4

Or:

If we write $k=n-1$ we get $$ (k+1)^{k}-1 = {k\choose 0}k^k+....{k\choose k-1}k$$ $$ = = {k\choose 0}k^k+....{k\choose k-2}k^2+ k^2 = k^2(......)$$

(since ${k\choose k-1} = k$)

nonuser
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