Proving ${p-1 \choose k}\equiv (-1)^{k}\pmod{p}: p \in \mathbb{P}$ 
Possible Duplicate:
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.
 A: It turns out that we do not even need Wilson's Theorem. Note the identity
$$\binom{p-1}{k+1}(k+1)=\binom{p-1}{k}(p-k-1).$$
This is easily obtained from the fact that $\binom{n}{m}=\frac{n!}{m!(n-m)!}$.
Now note that $p-k-1\equiv -(k+1)\pmod p$. Thus 
$$\binom{p-1}{k+1}(k+1)\equiv -\binom{p-1}{k}(k+1)\pmod p.$$
If $0\le k \lt p-1$, then $k+1$ is not divisible by $p$, so we can cancel and obtain
$$\binom{p-1}{k+1}\equiv -\binom{p-1}{k}\pmod p.\tag{$1$}$$
Thus $\binom{p-1}{k}$ changes sign modulo $p$ every time that we increment $k$. 
But $\binom{p-1}{0}=1$, and the result follows.
A: Remember Wilson's Theorem for a prime $\,p\,$: $$(p-1)!=-1\pmod p$$and from what you already proved we get $$\binom {p-1}{k}=\binom{p}{k}-\binom{p-1}{k-1}=\binom{p-1}{k-1}\pmod p$$Now just observe $$\frac{(p-1)!}{(p-k)!}=(p-k+1)(p-k+2)\cdot ...\cdot (p-1)\equiv(1-k)(2-k)\cdot ...\cdot (-1)(\text{mod } p)\Longrightarrow$$$$\Longrightarrow\binom{p-1}{k-1}=\frac{(1-k)(2-k)\cdot ...\cdot (-1)}{1\cdot 2\cdot ...\cdot (k-1)}=(-1)^k\pmod p $$
