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This problem is well-known, but my proof is different from the one I found, so just decided to put it here in case someone finds a mistake.

So I want to prove $p \mid m^p-m$, where $p$ is a prime number and $m \in \mathbb Z$. Obviously $m^p-m=m(m^{p-1}-1)$, and $p-1$ is an even number. Assuming $p \nmid m$, $m$ can be written as $p \pm 1, \pm 2, \ldots ,\pm \frac{p-1}{2},$ and then for $m=p \pm r, \ r \in \mathbb{Z}$, we have $p \mid m^{p-1}-1$.

I prove the conjecture by induction:

  1. $m=p+1$: The expression in the brackets becomes (expanding in Binomial series) $ (1+p)^{p-1}-1=\sum_{k=0}^{p-1}\binom{p-1}{k}p^k-1=\binom{p-1}{1}p+\binom{p-1}{2}p^2+ \cdots,$ since $1$ cancels out, and this is a multiple of $p$. If $m=p-1$: Applying the same idea, the expression in the brackets becomes $\sum\limits_{k=0}^{p-1}\binom{p-1}{k}p^k (-1)^{p-1-k}-1$, the first term is $(-1)^{p-1}$, and, since $p-1$ is even, it cancels out with $1$ and the rest is a multiple of $p$.

  2. Assume true for $m=p \pm r, \ r \in \mathbb Z$.

  3. $m=p \pm r \pm 1$: once again, expanding in binomial series, I obtain $\sum_{k=0}^{p-1}\binom{p-1}{k}(p+r)^k -1 .$ The $1$'s cancel out and the rest is a multiple of $(p+r)$, which, by Step 2 (assumption), $p \mid (p+r)^{p-1}-1$. Similar case when $p -r \pm 1$, which proves the conjecture.

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    @awllower: is it a good thing? But yes thanks, I will read that2011-09-07

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As mentioned in the comments, many proofs can be found here.

The purpose of this post is so that this question does not remain unanswered.