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If $x$ is a natural number and $p$ is a prime number, then $x^p - x$ is a multiple of $p$.

How would you go about proving this using induction?

I know that for any prime number $p$, and any integer $k$ such that $1 \leq k \leq p-1$, the binomial coefficient $\binom{p}{k}$ is divisible by $p$.

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    Now, use the expansion of $(x+1)^p$ for the induction step – 2017-01-30
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    Could you explain a bit more? – 2017-01-30
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    @Peter please write up a full answer. If your hints aren't enough, the OP will need it. If your hints are enough, this question will unnecessarily be marked "unanswered" forever (I'm not doing it myself because I don't want to take credit for your idea). – 2017-01-30

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Claim : If $x$ is a natural number and $p$ is a prime number , $x^p-x$ is divisible by $p$

Proof by induction over $x$ :

Base case : For $x=0$ and $x=1$ , the claim is true because $0$ is divisble by $p$. (If you considered $0$ to be natural, you can start with $x=0$, else with $x=1$)

Induction step : We assume that $x^p-x$ is divisble by $p$

Now, we have $$(x+1)^p=x^p+\binom{p}{1}x^{p-1}+\cdots+\binom{p}{p-1}x+1$$

This implies $$(x+1)^p-(x^p+1)=\binom{p}{1}x^{p-1}+\cdots+\binom{p}{p-1}x$$

hence $$(x+1)^p-(x^p+1)=(x+1)^p-x^p-1$$ is divisble by $p$ (The remark in your question comes into play here; the binomial coefficients are divisble by $p$, hence the right side is as well)

Now, we consider $$(x+1)^p-(x+1)=(x+1)^p-x-1=(x+1)^p-x^p-1+x^p-x$$

(We subtracted $x^p$ and added it again to get two summands divisble by $p$; Such tricks occur often in induction proofs)

We have already shown that $(x+1)^p-x^p-1$ is disivisble by $p$. By induction hypothesis, $x^p-x$ is divisble by $p$. Since $(x+1)^p-(x+1)$ is the sum of the two expressions, it is divisible by $p$ as well.

This completes the induction step and therefore the proof.