PROBLEM: provide a proof by contradiction of the following statement
"If $n,k \in \mathbb{N}$ and $\binom{n}{k}$ is a prime number, then $k=1$ or $k=n-1$."
My reasoning is shown below.
The statement can be translated in formal logic as shown below
P: $\forall n,k \in \mathbb{N}, (\binom{n}{k}$ is prime) $\Rightarrow (k=1 \wedge k=n-1)$
The negation is
not P: $\exists n,k \in \mathbb{N}, (\binom{n}{k}$ is prime)$ \wedge (k\neq1) \wedge (k\neq n-1)$
The goal is to prove that not P leads to a contradiction. Let's consider some trivial cases to narrow down the problem.
Case $k=n \rightarrow \binom{n}{k}=\binom{n}{n}=1$ for all $n \in \mathbb{N}$, but $1$ is not prime, so we have a contradiction.
Case $k>n \rightarrow \binom{n}{k}=0$ for all $n\in \mathbb{N}$, but $0$ is not prime, so we have a contradiction.
So from $(k\neq1) \wedge (k\neq n-1)$ above and from these two cases we conclude that $1<k<n-1$, or equivalently $2\leq k \leq n-2$. So our statement not P becomes
not P: $\exists n,k \in \mathbb{N}, (\binom{n}{k}$ is prime)$ \wedge (2\leq k \leq n-2)$
and expanding the definition of binomial coefficient, and calling $p$ a prime
not P: $\exists n,k \in \mathbb{N}, 2\leq k \leq n-2, \frac{n\cdot (n-1)\cdot(n-2) \cdots (n-k+1) }{k!}=p$
any ideas to continue?
EDIT: this post (link) provides a direct proof, but I am looking for a proof by contradiction, so my post is not a duplicate.