Binomial Coefficient of Prime Minus One Modulo Prime

Theorem

Let $p$ be a prime number.

Then:

$0 \le k \le p - 1 \implies \dbinom {p - 1} k \equiv \left({-1}\right)^k \pmod p$

where $\dbinom {p - 1} k$ denotes a binomial coefficient.

$\dbinom p k \equiv 0 \pmod p$

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

$\dbinom {p-1} k + \dbinom {p - 1} {k - 1} = \dbinom p k \equiv 0 \pmod p$

This certainly holds for $k = 1$, and so we have:

$\dbinom {p - 1} 1 + \dbinom {p - 1} 0 = \dbinom p 1 \equiv 0 \pmod p$

$\dbinom {p - 1} 0 = 1 \equiv 1 \pmod p$

So:

$\dbinom {p - 1} 1 \equiv -1 \pmod p$

Then:

$\dbinom {p - 1} 2 + \dbinom {p - 1} 1 = \dbinom p 2 \equiv 0 \pmod p$

and so:

$\dbinom {p - 1} 2 \equiv 1 \pmod p$

The result follows.

$\blacksquare$

1991: Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery: An Introduction to the Theory of Numbers (5th ed.): $\S 2.1$: Exercise $44$
1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $10 \ \text{(c)}$