TITLE: how compute $\gcd\Bigl(\binom{n}{1},\binom{n}{2},\binom{n}{3},...,\binom{n}{n-1}\Bigr)?$
QUESTION [5 upvotes]: Let $n > 1$ be a positive integer. Prove that
$$\gcd\Biggl(\binom{n}{1},\binom{n}{2},\binom{n}{3},...,\binom{n}{n-1}\Biggr)=\begin{cases}
p,  & \text{if $n=p^m$ with $p$ being prime} \\
1, & \text{otherwise}  \\
\end{cases}$$

REPLY [3 votes]: HINT: If $p$ is prime, do you know that $n!$ is divisible by $p^r$, but not by $p^{r+1}$, where $r=r(n)=\lfloor \frac np\rfloor+\lfloor\frac n{p^2}\rfloor +\lfloor\frac n{p^3}\rfloor +\ldots$?
With this one sees that the exponent of $p$ in $n\choose k$ is $r(n)-r(k)-r(n-k)$