Binomial Coefficient of Prime Minus One Modulo Prime

Theorem
Let $$p$$ be a prime number.

Then:
 * $$0 \le k \le p-1 \implies \binom {p-1} k \equiv \left({-1}\right)^k \pmod p$$

where $$\binom {p-1} k$$ is a binomial coefficient.

Proof
From Binomial Coefficient of Prime, we have:
 * $$\binom p k \equiv 0 \pmod p$$

when $$1 \le k \le p-1$$.

From Pascal's Rule we have:
 * $$\binom {p-1} k + \binom {p-1} {k - 1} = \binom p k \equiv 0 \pmod p$$.

This certainly holds for $$k = 1$$, and so we have:
 * $$\binom {p-1} 1 + \binom {p-1} 0 = \binom p 1 \equiv 0 \pmod p$$

But $$\binom {p-1} 0 = 1 \equiv 1 \pmod p$$.

So $$\binom {p-1} 1 \equiv -1 \pmod p$$.

Then:
 * $$\binom {p-1} 2 + \binom {p-1} 1 = \binom p 2 \equiv 0 \pmod p$$

... and so $$\binom {p-1} 2 \equiv 1 \pmod p$$.

The result follows.