The Greatest Common Divisor of a Binomial Row

Write $h(n) = \\gcd\\big(\\binom{n}{1}, \\dots, \\binom{n}{n-1}\\big)$. The answer is governed entirely by carries in base-$p$ addition: by Kummer's theorem the exponent of $p$ in $\\binom{n}{i}$ counts the carries when $i$ and $n-i$ are added in base $p$.From that criterion, $h(n) = p$ when $n$ is a power of the prime $p$, and $h(n) = 1$ otherwise. The intermediate observation worth isolating is that divisibility at a single index proves nothing: whenever $p \\mid n$ the index $i = p-1$ forces a carry out of the units place, so $p \\mid \\binom{n}{p-1}$ always. Only a prime dividing the \\emph{whole} row contributes to the gcd.