Is it true that for integers $n > e, x \ge n$, it is impossible for $\dfrac{(x+n)!}{x!}$ to be divisible by $n$ distinct primes where each prime is greater than $\dfrac{(x+n)e}{n}$?
Example:
- $x=n=10$
There cannot be $10$ primes greater than $\dfrac{20}{e}{10} \approx 5.44$ that divide $$\dfrac{20!}{10!} = 11\times 12\times 13\times 14\times 15\times 16\times 17\times 18\times 19\times 20$$
It is not possible for $7^{10}$ or any combination of $10$ primes where each prime is greater or equal to $7$ to divide $\dfrac{20!}{10!}$
Here's my thinking:
(1) Let $n>1$ and $x \ge n$ be integers.
(2) If a prime $p>n$ divides $\dfrac{(x+n)!}{x!}$, then it necessarily divides ${{x+n}\choose{n}}$ since it is not divisible by $n!$.
(3) From a well known inequality:
$${{x+n}\choose{n}} < \left(\frac{(x+n)e}{n}\right)^n$$
(4) If all $n$ primes are greater than $\dfrac{(x+n)e}{n}$, then it follows:
$$p_1{p_2}\dots{p_n} > \left(\frac{(x+n)e}{n}\right)^n > {{x+n}\choose{n}}$$
