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Prove $\binom{p-1}{k} \equiv (-1)^k\pmod p$
The question is as follows:
Let $p$ be prime. Show that ${p \choose k}\bmod{p}=0$, for $0 \lt k \lt p,\space k\in\mathbb{N}$. What does this imply about the binomial co-efficients ${p-1 \choose k}$?
By the definition of binomial coefficients:
${p \choose k}=\frac{p!}{k!(p-k)!}$
Now if $0 \lt k \lt p$, then we have $p\mid{p\choose k}$, therefore ${p \choose k}\equiv0\pmod{p}, \space 0 \lt k \lt p. \space \blacksquare$
Note that we can write: ${p \choose k}={p-1 \choose k}+{p-1 \choose k-1}$, and therefore:
${p-1 \choose k}={p \choose k}-{p-1 \choose k-1}=\frac{p!}{k!(p-k)!}-\frac{(p-1)!}{(k-1)!(p-k)!}=\frac{(p-1)!}{(k-1)!(p-k)!}\left(\frac{p}{k}-1\right)$
However, I am unsure how to proceed with this question, the book I am working from states that:
${p-1 \choose k}\equiv(-1)^{k}\pmod{p}, \space 0 \le k \lt p$
But I am unsure how the authors have derived this congruence, so I'd appreciate any hints.
Thanks in advance.