By the binomial theorem, we have $(xa)^n = \sum_{k=0}^n \binom{n}{k} x^k a^{n-k}$. We want to show that for each $k$, $1 \leq k \leq n-1$, the term $\binom{n}{k} x^k a^{n-k}$ is congruent to $0 \pmod{n}$. Since $n$ is prime, we know that $\binom{n}{k} \equiv 0
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