Prove that $\gcd{\left(\binom M1,\binom M2,\binom M3,\ldots,\binom Mn\right)}=1$ where $M=\mathrm{lcm}(1,2,3,\ldots,n)$
HINT: Suppose that $p$ is a prime dividing $M$, and that $p^k$ is the highest power of $p$ dividing $M$.
- Show that $p^k\le n$.
- Show that if $M=ap^k$, and $1\le t\le p^k$, then $(a-1)p^k+t$ is divisible by the same maximum power of $p$ as $t$.
- Conclude that $p$ does not divide $\dbinom{M}{p^k}$.
Hint: Apply Lucas' Theorem which states that for non-negative integers $m$ and $n$, and a prime $p$, $$ {m \choose n} \equiv \prod_{i=0}^{k} {m_{i} \choose n_{i}} \pmod p,$$ where $m=m_{k}p^{k}+m_{k-1}p^{k-1}+\cdots+m_{1}p+m_{0}$ and $n=n_{k}p^{k}+n_{k-1}p^{k-1}+\cdots+n_{1}p+n_{0}$ are the base $p$ expansions of $m$ and $n,$ respectively. This uses the convention that ${m \choose n} =0$ when $m<n$ and ${0\choose 0 } = 1$.
Corollary: For any $M$, if $n \geq p^k$ for every prime $p$ such that $M = p^k q$ where $q \not\mid p$, then
$$\gcd{\left(\binom{M}{1},\binom{M}{2},\binom{M}{3},\ldots,\binom{M}{n}\right)}=1$$
Proof: Fix prime $p$, let $M=p^k q$ where $ q\not \mid p$. Then from Lucas' Theorem, ${M \choose p^k} \neq 0 \pmod{p}$. (In fact, it is equal to $q \pmod{p}$.)
Hence, the gcd is not divisible by any prime, so must be equal to 1.
The condition in the corollary is clearly satisfied when $M=\mathrm{lcm}(1,2,3,\ldots,n).$