Binomial Coefficient of Prime/Proof 3

Theorem
Let $p$ be a prime number.

Then:
 * $\displaystyle \forall k \in \Z: 0 < k < p: \binom p k \equiv 0 \pmod p$

where $\displaystyle \binom p k$ is defined as a binomial coefficient.

Proof
By the definition of binomial coefficient:

Now, $p$ divides the LHS by Factorial Divides All Numbers Less than It.

So $p$ must also divide the RHS.

By hypothesis $k < p$.

Writing:


 * $k! = k\left({k-1}\right)\ldots\left({2}\right)\left({1}\right)$

we see that $p$ is not a factor of $k!$, as $p$ is prime, and each factor is less than $p$.

Similarly, if we write:


 * $\left({p-k}\right)! = \left({p-k}\right)\left({p-k-1}\right)\ldots\left({2}\right)\left({1}\right)$

it follows that $p$ is also not a factor of $\left({p-k}\right)!$, by the same reasoning as above.

The result then follows from Euclid's Lemma.