How Does Modulo Arithmetic Prove a Combinatorial Identity Involving Primes?

[evinda]

Hey again! (Blush)

If $p \neq 2$ is a prime and $0 \leq k \leq p-1, \text{ prove that: }$
$$\binom{p-1}{k} \equiv (-1)^k \pmod p$$

I thought that I could calculate $\binom{p-1}{k}k!$ :

$$\binom{p-1}{k}k!=\frac{(p-1)!}{k!(p-(k+1))!}k!=\frac{(p-1)!}{(p-(k+1))!}$$

According to Wilson's theorem, $(p-1)! \equiv -1 \pmod p$..
But...what is equal $\mod p$ to $(p-(k+1))!$ ? (Thinking)

 

[evinda]

Hey again! (Blush)

If $p \neq 2$ is a prime and $0 \leq k \leq p-1, \text{ prove that: }$
$$\binom{p-1}{k} \equiv (-1)^k \pmod p$$

I thought that I could calculate $\binom{p-1}{k}k!$ :

$$\binom{p-1}{k}k!=\frac{(p-1)!}{k!(p-(k+1))!}k!=\frac{(p-1)!}{(p-(k+1))!}$$

According to Wilson's theorem, $(p-1)! \equiv -1 \pmod p$..
But...what is equal $\mod p$ to $(p-(k+1))!$ ? (Thinking)

Could I show it maybe like that?

$$ \binom{p-1}{k}=\frac{(p-1)!}{k!(p-(k+1))!}=\frac{(p-k)(p-k+1) \cdots (p-1)}{k!} \equiv \frac{(p-1) \cdots (p-k+1)(p-k)}{k!} \equiv \frac{(-1) \cdots (-k+1)(-k)}{k!} \equiv \frac{(-1)^k k!}{k!} \equiv (-1)^k \pmod p $$ ?
(Nerd)

 

[I like Serena]

Could I show it maybe like that?

$$ \binom{p-1}{k}=\frac{(p-1)!}{k!(p-(k+1))!}=\frac{(p-k)(p-k+1) \cdots (p-1)}{k!} \equiv \frac{(p-1) \cdots (p-k+1)(p-k)}{k!} \equiv \frac{(-1) \cdots (-k+1)(-k)}{k!} \equiv \frac{(-1)^k k!}{k!} \equiv (-1)^k \pmod p $$ ?
(Nerd)

Looks good! (Nod)

 

[evinda]

Looks good! (Nod)

Nice,thank you very much! (Happy)