Is (p-1)C(a) congruent to (-1)^a mod(p)?

[Poirot1]

Let p be prime and a be between 1 and p-1. Show the binomial coefficient (p-1)C(a) satifies

(p-1)C(a) =(-1)^a mod(p).

(p-1)C(a) =$\frac{(p-1)!}{a!(p-1-a)!}$ so we can apply wilson's theorem which says
(p-1)!=-1 (modp)

 

[caffeinemachine]

Re: congruence equation

Let p be prime and a be between 1 and p-1. Show the binomial coefficient (p-1)C(a) satifies

(p-1)C(a) =(-1)^a mod(p).

(p-1)C(a) =$\frac{(p-1)!}{a!(p-1-a)!}$ so we can apply wilson's theorem which says
(p-1)!=-1 (modp)

Let $x\binom{p-1}{a+1}\equiv \binom{p-1}{a}\pmod{p}$.

Then we have $\frac{x(p-1)!}{(p-a-2)!(a+1)!}\equiv \frac{(p-1)!}{(p-a-1)!a!}\pmod{p}$.

Cancel things out (why can that be done?), you get $x\equiv -1\pmod{p}$

Now apply induction.