Least Common Multiple and the product of a sequence of consecutive integers Let $x>0, n>0$ be integers.
Let LCM$(x+1, x+2, \dots, x+n)$ be the least common multiple of $x+1, x+2, \dots, x+n$.
Let $v_p(u)$ be the highest power of $p$ that divides $u$.
It seems to me that:
$$\frac{(x+1)(x+2)\times\dots\times(x+n)}{\text{LCM}(x+1, x+2, \dots, x+n)} \le (n-1)!$$
Here is my thinking:


*

*Let $p$ be any prime that divides $(x+1)\times\dots\times(x+n)$.

*Let $x+i$ be the integer in $\{x+1,x+2. \dots, x+n\}$ that is divisible by the highest power of $p$.

*All integers in the sequence $\{x+1, x+2, x+3, \dots, x+n\}$ that are divisible by $p$ will have the form $x + i \pm ap$ where $0 < a \le \left\lfloor\dfrac{n-1}{p}\right\rfloor$.

*$v_p\left(\dfrac{(x+1)(x+2)\times\dots\times(x+n)}{\text{LCM}(x+1,x+2,\dots,x+n)}\right) \le v_p(p) + v_p(2p) + \dots + v_p\left(\left\lfloor\dfrac{n-1}{p}\right\rfloor p\right) = v_p((n-1)!)$

*So, it follows that the maximum power of $p$ that divides $\dfrac{(x+1)\times\dots\times(x+n)}{\text{LCM}(x+1,\dots,x+n)}$ is less than or equal to the maximum power of $p$ that divides $(n-1)!$

Edit:  Made changes to third and fourth bullet point with an attempt to make them cleaner.
Based on feedback received.

Edit 2:  I originally put the "same" power in the last bullet point.  That is not correct.  It should be less or equal.  I have changed it.

Edit 3:  Fixed a typo.  n! should be (n-1)!.

Your claim is true according to a wiki page.
Changing $n,k$ to $x+n,n-1$ respectively in$$\binom nk\le\frac{\text{LCM}(n-k,n-k+1,\cdots, n)}{n-k}$$
we get
$$\binom{x+n}{n-1}\le\frac{\text{LCM}(x+1,x+2,\cdots, x+n)}{x+1}$$
which is your claim.
