Applying Wilson's Theorem

Question

Prove the following: $\frac{(p-1)!}{(p-1-a)! a!}= (-1)^a \mod p \ $ with $0 \leq a \leq p-1$ and $p$ being an odd prime.

I know that $(p-1)!= -1 \mod p \ $ by Wilson's Theorem but I'm unable to complete the proof.

Answer 1

Let $$\frac{(p-1)!}{(p-1-a)!a!}=N.\tag{1}$$ Then $N$ is an integer, for it is the number of ways to choose $a$ objects from $p-1$ objects. In number theory, working with "fractions" can be dangerous, so we rewrite (1) as $$(p-1)!=(p-1-a)!a!N.\tag{2}$$ Note that $a\equiv -(p-a)\pmod{p}$ and $a-1\equiv -(p-a+1)\pmod{p}$ and so on down to $1\equiv -(p-1)\pmod{p}$.

Thus $(p-1-a)!a!$ is congruent to $(-1)^a(p-1)!$ modulo $p$. We conclude that $$(p-1)!\equiv (-1)^a(p-1)!N\pmod{p}.$$ But $(p-1)!$ is relatively prime to $p$. Thus by cancellation we have $N\equiv (-1)^a\pmod{p}$.

Answer 2

$$ (p-1-a)! a! \equiv_p \prod_{k=1+a}^{p-1}(p-k) \prod_{j=1}^a j \equiv_p (-1)^{p-1-a}\prod_{k=1+a}^{p-1}k \prod_{j=1}^a j\equiv_p (-1)^{a}(p-1)! $$