Prove that $\operatorname{lcm}$ of [$\binom{n}{1}$, $\binom{n}{2}$, ... ,$\binom{n}{n}$] = $\operatorname{lcm}(1, 2, ...,n+1)/(n+1)$ How to prove that:
$$\operatorname{lcm}\left(\binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n}\right) = \frac{\operatorname{lcm}(1, 2, \ldots, n+1)}{n+1}$$

There is a hint: 
$p$ is a prime and consider the highest power of $p$ in $\binom{n}{k}$, that is $p^e \:||\: \binom{n}{k}$, e is the highest power of p.

Lemma: $[\frac{n}{p}]-[\frac{n-k}{p}]-[\frac{k}{p}]\le 1$
Note that $$v_p(\binom{n}{k})=\sum_{i=1}^{\infty} [\frac{n}{p^i}]-[\frac{n-k}{p^i}]-[\frac{k}{p^i}]$$
Suppose $p^e\le n+1 <p^{e+1}$ for some $p<n$, then $v_p(\operatorname{lcm}(1, ..., n+1))=e$.
We need to prove that
$$(n+1)v_p(\operatorname{lcm}(\binom{n}{1}, ..., \binom{n}{n}) = v_p((n+1)\operatorname{lcm}(\binom{n}{1}, ..., \binom{n}{n}) = e$$
Now $(n+1)\binom{n}{k} = (k+1)\binom{n+1}{k+1}$, so $$v_p((n+1)\operatorname{lcm}(\binom{n}{1}, ..., \binom{n}{n}) = \max\limits_{1\le k \le n}v_p((k+1)\binom{n+1}{k+1})=\max\limits_{1\le k\le n}(v_p(k+1)+v_p\binom{n+1}{k+1})$$
Let $v_p(k+1)=s$ then by the lemma $v_p(\binom{n+1}{k+1}) \le e-s$, therefore $v_p((k+1)\binom{n+1}{k+1}) \le e$ for all $k=1, 2, ..., n$.
Also if $k=p^e-1$, $v_p((k+1)\binom{n+1}{k+1}) \ge e$. So we are done.
