Binomial divisibility

The problem says to ignore the first and last terms, so assume $n \geq 2$ so there are at least $3$ terms in the expansion, and $1 \leq k \leq (n-1).$ The coefficient of the term $x^k$ in the expansion of $(1+x)^n$ is the binomial coefficient $c_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}.$ $c_k$ is always an integer since $k!$ and $(n-k)!$ are factors of $n!,$ and if $n$ is prime, then $n=p,\ p\in\mathbb{P}.$ $c_k = \frac{1\times 2\times 3\times ...\times p}{(1\times 2\times ...\times k)(1\times 2\times ...\times (p-k))}$ Assumed above, $1 \leq k \leq (n-1).$ Therefore $1\leq k\leq (p-1)$ and $1\leq (p-k)\leq (p-1).$ Since both terms in the denominator are less than $p$ and $p$ cannot be written as the product of positive integers less than itself, $p$ is not a factor of the denominator. Therefore the quotient above is a multiple of $p,$ or $p$ itself.

On the other hand, in the case where $n$ is a composite number $m,$ $c_k = \frac{1\times 2\times 3\times ...\times m}{(1\times 2\times ...\times k)(1\times 2\times ...\times (m-k))}$ There must be at least one $k$ (and its opposite, $(m-k),$ ) such that the quotient above is not a multiple of $m$ or $m$ itself, but a multiple of a smaller factor of $m.$ Therefore when $n$ is composite, there are at the very least $2$ coefficients $c_k$ that are not multiples of $n$ in the expansion.

Therefore, when $n\geq 2,$ every coefficient except those of the first and last terms in the expansion of $(1+x)^n$ is divisible by $n$ if and only if $n\in \mathbb{P}.$

Observation of Pascal Triangle shows that it is not odd or even nor square number. However 2,3,5, 7 works. Hence the answer after checking for 11 and 13. Yes, This is NOT a proof. I agree.