Since $ n! = \prod_{p_{i}\text{ prime}}p_{i}^{\sum^{\lfloor\log_{p_{i}}(n)\rfloor}_{k = 1}\left\lfloor\frac {n}{p_{i}^k}\right\rfloor}$ and $ \text{lcm}(1,\ldots,n) = \prod_{p_{i}\text{ prime}}p_{i}^{\lfloor\log_{p_{i}}(n)\rfloor}$, there is to prove that \[ \binom{2n}{n} = \prod_{p_{i}\text{ prime}}p_{i}^{\sum^{\lfloor\log_{p_{i}}(2n)\rfloor}_{k = 1}\left\lfloor\frac {2n}{p_{i}^k}\right\rfloor - 2\left\lfloor\frac {n}{p_{i}^k}\right\rfloor}\left|\prod_{p_{i}\text{ is prime}}p_{i}^{\lfloor\log_{p_{i}}(2n)\rfloor}\right. = \text{lcm}(1,\ldots,2n),\] so it suffices to show that $ \sum^{\lfloor\log_{p_{i}}(2n)\rfloor}_{k = 1}\left\lfloor\frac {2n}{p_{i}^k}\right\rfloor - 2\left\lfloor\frac {n}{p_{i}^k}\right\rfloor\le\lfloor\log_{p_{i}}(2n)\rfloor$, which is obvious since $ \left\lfloor\frac {2n}{p_{i}^k}\right\rfloor - 2\left\lfloor\frac {n}{p_{i}^k}\right\rfloor\in\{0,1\}$. [Moderator edit: fixed some typos. - darij]

Peter wrote: Since $ n! = \prod_{p_{i}\text{ prime}}p_{i}^{\sum^{\lfloor\log_{p_{i}}(n)\rfloor}_{k = 1}\left\lfloor\frac {n}{p_{i}^k}\right\rfloor}$ There's no need to know this... a more general statement holds: for every $ k < 2n$: $ 2n{2n - 1\choose k}\mid \text{lcm} (2n,2n - 1,\cdots ,2n - k)$ (note that by letting $ k = n$ we will derive the first problem) Let $ p^{\alpha}$ be the greatest power of $ p$ which divides $ \text{lcm} (2n,2n - 1,\cdots ,2n - k)$.hence there exists a natural number $ t$,in which $ p^{\alpha}\mid 2n - t$,therefore for every $ i\leq\alpha$ we would have $ p^i\mid 2n - t$ or in other words $ 2n + t - l\equiv l\pmod{p^i}$,($ 1\leq l\leq t$) thus exactly $ \lfloor\frac {t}{p^i}\rfloor$ numbers from the set $ \{ 2n - t - 1,\cdots ,2n\}$ and $ \lfloor\frac {k - t}{p^i}\rfloor$ numbers from the set $ \{2n - t + 1,\cdots , 2n - k\}$ are divisible by $ p^i$.on the other hand,we know that $ \lfloor\frac {t}{p^i}\rfloor + \lfloor\frac {k - t}{p^i}\rfloor\leq\lfloor\frac {k}{p^i}\rfloor$ which means that the number of multiples of $ p^i$ in the two sets mentioned above are less than the number of multiples of $ p^i$ in the set $ \{1,2,\cdots ,k\}$. thus the greatest power of $ p$ that divides $ 2n {2n - 1\choose k} = \frac {2n(2n - 1)\cdots (2n - t + 1)(2n - t - 1)\cdots (2n - k)(2n - t)}{k!}$ is less than or equal to $ \alpha$,hence: $ 2n{2n - 1\choose k}\mid\text{lcm} (2n,2n - 1,\cdots ,2n - k)$

Just wondering, but does this solution work? : Solution (My-don't-know-if-its-correct try) : Lemma: If $x\ge0$ and is real number, then we have \[\ 2[x]\le [2x] \le 2[x]+1 \Rightarrow 0 \le [2x]-2[x] \le 1\] Proof: If $x$ is an integer, then clearly, we have $2[x]=[2x]$. When $x$ is real, then let $x=[x]+\{x\}$, assume that $\{x\} < 0.5$ then, $[2x]=[2[x]+2{x}]=2[x]$. If $\{x\} \ge 0.5$ then we have $[2x]=[2[x]+2{x}]=2[x]+1>2[x]$ $\Box$. Let $p$ be a prime, such that $p|\dbinom{2n}{n}$ hence, $p|lcm(1,...,2n)$. Let $k$ be the greatest integer such that $p^k \le 2n <p^{k+1}$. Now we apply Legenrde's Function, we have (as the exponent of $p$) \[\ \sum_{i=0}^{\infty} \lfloor \frac{2n}{p^i} \rfloor -2 \sum_{i=0}^{\infty} \lfloor \frac{n}{p^i} \rfloor \le k \] Our aim now is to prove that the above inequality is true. And note that the RHS is equal to $k$ is because of our definition. Now we note that when $i>k$ the LHS will not produce any more positive values, for all $i \le k$, we can apply the lemma, which gives us, \[\ 0 \le \sum_{i=0}^{\infty} \lfloor \frac{2n}{p^i} \rfloor -2 \sum_{i=0}^{\infty} \lfloor \frac{n}{p^i} \rfloor \le k(1) \] as desired $\Box$ EDIT: some typos

Peter wrote: ... which is obvious since $ \left\lfloor\frac {2n}{p_{i}^k}\right\rfloor - 2\left\lfloor\frac {n}{p_{i}^k}\right\rfloor\in\{0,1\}$. Is this not exactly the crux of your proof? What new do you bring to Peter's ?

OMG Sorry, I thought we had completely different solutions due to his "complex notation" (i just typed up my solution right after i thought of it )