GCD of binomial coefficients Can the following be proved or disproved?
$$\gcd\left(\binom{n}{1} , \binom{n}{2} , \binom{n}{3},...........,\binom{n}{\lfloor \frac n2 \rfloor}\right)$$
Where $n \ge 4$ and is a positive integer
 Is always a prime number or 1.
It would be very helpful if the way to prove or disprove it may make use of the properties of Pascal's triangle.
 A: $hint$
if n is a prime then $n\choose r$ is always a multiple of $n$ because non of factor of denominator can cancel $n$ 
A: Kummer's Theorem
As shown in this answer, Kummer's Theorem says that the number of factors of a prime $p$ that divide $\binom{n}{k}$ is the number of carries that are performed when adding $k$ and $n-k$ in base-$p$.

If $\boldsymbol{n\ne p^m}$, then there is a $\boldsymbol{k\not\in\{0,n\}}$ so that $\boldsymbol{p\nmid\binom{n}{k}}$
If $n\ne p^m$ for some prime $p$ and integer $m\ge0$, then, in base-$p$, either the highest order digit is not $1$ or there is a lower order non-zero digit.

In the first case,
$$
n=\sum_{j=0}^md_jp^j
$$
where $d_m\gt1$, so we can take
$$
\begin{align}
k
&=n-p^m\\
&=\sum_{j=0}^m(d_j-[j=m])p^j\\
&\ge(d_m-1)p^m
\end{align}
$$
so that $k\not\in\{0,n\}$ and  $k+p^m=n$ with no carry, i.e. $(d_m-1)+1=d_m$,  means that $p\nmid\binom{n}{k}$.

In the second case,
$$
n=\sum_{j=0}^md_jp^j
$$
where there is some $j_0\lt m$ so that $d_{j_0}\gt0$, so we can take
$$
\begin{align}
k
&=n-p^{j_0}\\
&=\sum_{j=0}^m(d_j-[j=j_0])p^j\\
&\ge d_mp^m
\end{align}
$$
so that $k\not\in\{0,n\}$ and  $k+p^{j_0}=n$ with no carry, i.e. $(d_{j_0}-1)+1=d_{j_0}$,  means that $p\nmid\binom{n}{k}$.

If $\boldsymbol{n=p^m}$, then for all $\boldsymbol{k\not\in\{0,n\}}$, we have $\boldsymbol{p\mid\binom{n}{k}}$
Subtracting any $k\not\in\{0,n\}$ will require a borrow from the $1$ in the $p^m$ place, so there will be a corresponding carry when adding $k$ to $n-k$. Thus, $p\mid\binom{n}{k}$.
Furthermore, there is one base-$p$ carry when adding $(p-1)p^{m-1}+p^{m-1}=p^m$, so there is only one factor of $p$ in $\binom{p^m}{p^{m-1}}$.

Conclusion
Thus,
$$
\gcd\left[\binom{n}{k}\right]_{k=1}^{n-1}=\left\{\begin{array}{}
p&\text{if }n=p^m\text{ for some prime $p$ and integer $m\gt0$}\\
1&\text{if }n\ne p^m\text{ for any prime $p$ and integer $m\gt0$}\end{array}\right.
$$
A: $\newcommand{\Z}{\mathbb{Z}}$Suppose $n$ is not a prime-power, and let 
$$n = \prod_{i=1}^{k} p_{i}^{e_{i}}$$ 
with $p_{i}$ distinct primes, $e_{i} > 0$ for each $i$.
Write
$$n_{j} = \prod_{i \ne j} p_{i}^{e_{i}}.$$
Then it is well known that
$$
\binom{n}{p_{i}^{e_{i}}} \equiv \binom{p_{i}^{e_{i}} n_{i}}{p_{i}^{e_{i}}} \equiv n_{i} \pmod{p},
$$
so that $\dbinom{n}{p_{i}^{e_{i}}}$ is not divisible by $p_{i}$. Since your gcd is a divisor of $n = \dbinom{n}{1}$, this shows that the gcd is $1$ in this case.
If $n = p^{e}$ is a power of the prime $p$, then the equation in $\Z/p\Z[x]$
$$
(1 + x)^{p^{e}} = 1 + x^{p^{e}}
$$
shows that all of your binomial coefficients are divisible by $p$, and their gcd divides $p^e = \dbinom{p^e} {1}$.
But by Kummer, the highest power of $p$ dividing 
$$
\binom{p^{e}}{p^{e-1}}
$$
is $p$, as there is precisely one carry in the $p$-adic addition of $p^{e-1}$ and $p^{e} - p^{e-1} = (p-1) p^{e-1}$, so that your gcd is $p$.
