Proof $\binom{p-1}{k}\equiv (-1)^k \mod{p}$

Question

Proof that if $p$ is a prime odd and $k$ is a integer such that $1≤k≤p-1$ , then the binomial coefficient

$$\displaystyle \binom{p-1}{k}\equiv (-1)^k \mod p$$

This exercise was on a test and I could not do!!

Answer 1

Let $a=\binom{p-1}{k}$. Then $$a k!=(p-1)(p-2)(p-3)\cdots (p-k).$$ The $i$-th term on the right-hand side is congruent to $-i$ modulo $p$. Thus $$ak!\equiv (-1)^k k!\pmod{p}.$$ Now since $k!$ is not divisible by $p$ we can cancel.

Answer 2

You can prove it by induction on $k$. If $ k=1$ $\to$ $\displaystyle \binom{p-1}{k} = p-1$ that $p-1 \equiv -1 \mod p$. For $k= n +1$ use this $\displaystyle \binom{m}{n+1} =\displaystyle \binom{m}{n} +\displaystyle \binom{m}{n+1}$

Answer 3

$$\binom{p-1}k=\frac{(p-1)\cdots(p-k)}{k!}=\prod_{r=1}^k\frac{p-r}r$$

Now $p-r\equiv-r\pmod p\implies\dfrac{p-r}r\equiv-1$

Answer 4

In characteristic $p$, where $p$ is odd,

$$(1+X)^{p-1} = \frac{(1+X)^p}{1+X} = \frac{1+X^p}{1+X} = \frac{1-(-X)^p}{1-(-X)} = 1 -X + X^2 - \dots +X^{p-1}.$$