Non-Divisbility of Binomial Coefficients of n by Prime

Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.

Let $p$ be a prime number.

Then:
 * $\dbinom n k$ is not divisible by $p$ for any $k \in \Z_{\ge 0}$ where $0 \le k \le n$


 * $n = a p^m - 1$ where $1 \le a < p$
 * $n = a p^m - 1$ where $1 \le a < p$

for some $m \in \Z_{\ge 0}$.

Proof
The statement:
 * $\dbinom n k$ is not divisible by $p$

is equivalent to:
 * $\dbinom n k \not \equiv 0 \pmod p$

The corollary to Lucas' Theorem gives:
 * $\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$

where:
 * $n, k \in \Z_{\ge 0}$ and $p$ is prime
 * the representations of $n$ and $k$ to the base $p$ are given by:
 * $n = a_r p^r + \cdots + a_1 p + a_0$
 * $k = b_r p^r + \cdots + b_1 p + b_0$

Consider the form of $n = a p^m - 1$ when represented to the base $p$:

That is, all of the digits of $n$ are $p - 1$ except perhaps the first one.

For $k$ such that $0 \le k \le n$, the digits of $k$ will all range over $0$ to $p - 1$.

Necessary Condition
Let $n = a p^m - 1$.

Then all the binomial coefficients $\dbinom {a_j} {b_j}$ (except perhaps the first) are of the form $\dbinom {p - 1} k$ for $0 \le k < p$.

By Binomial Coefficient of Prime Minus One Modulo Prime:
 * $\dbinom {p - 1} k \equiv \paren {-1}^k \pmod p$

and so:
 * $\dbinom {a_j} {b_j} \equiv \paren {-1}^k \pmod p$ for $0 \le j < r$

For the first digit, we have $0 \le b_r \le a_r < p$.

We have $\dbinom {a_r} {b_r} > 0$.

Since $a_r!$ is not divisible by $p$, neither does $\dbinom {a_r} {b_r}$.

Thus:
 * $\dbinom {a_r} {b_r} \not \equiv 0 \pmod p$

Hence:
 * $\ds \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \equiv \pm \dbinom {a_r} {b_r} \not \equiv 0 \pmod p$

which means $\dbinom n k$ is not divisible by $p$ for any $k \in \Z_{\ge 0}$ where $0 \le k \le n$.

Sufficient Condition
We show the contrapositive.

Suppose $n \ne a p^m - 1$.

Then one of the digits $a_j$ (that is not the first one) will be less than $p - 1$.

But the digits of all possible $k$ will range over $0$ to $p - 1$.

Therefore there must be at least one $\dbinom {a_j} {b_j}$ such that $b_j > a_j$.

Such a condition leads to $\dbinom {a_j} {b_j} = 0$.

That is:
 * $\ds \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \equiv 0 \pmod p$

which means $\dbinom n k$ is divisible by $p$ for some $k \le n$.