If $p=2$ then $\gcd( \binom{8}{1},..., \binom{8}{7})= 8$ If $p>2$ $\binom{2^pp}{1}=2^p *p$ For odd $n \neq k$ and $n \leq 2^p p-1 $ we have that $n\not | 2^p* p $ so $\frac{\binom{2^pp}{n}}{2^p* p}$ is integer So enough to find $\gcd( 2^p p, \binom{2^pp}{p})$ and easy to show that $2^p | \binom{2^pp}{p}$ $v_p( \binom{2^pp}{p}) =v_p( \frac{(2^p p)(2^p p-1)...(2^p p-p+1)}{p!})=0$ So answer is $2^p$

What does k mean in this solution?