Let $n \ge 2$ be an integer and $p$ be a prime that divides $n!$.
Let $x \ge n$ be an integer.
Let $v_p(u)$ be the highest power of $p$ that divides $u$.
It is well known that $n!$ divides $\dfrac{(x+n)!}{x!}$.
Let $x+j$ be the integer in the sequence $\{x+1, x+2, \dots, x+n\}$ divisible by the highest power of $p$.
Does it follow that $v_p(x+j) \ge v_p({{x+n}\choose{n}})$?
I am thinking that the answer is yes.
Here's my thinking:
(1) Let $p^t$ be the highest power of $p$ in the sequence ${x+1, x+2, \dots, x+n}$.
(2) Using Legendre's Formula, it is well known that $p^{v_p({{x+n}\choose{n}})} \le x+n$ [See see this paper for the argument].
(3) So, $v_p({{x+n} \choose {x}}) \le t$ and $p^{v_p\left({{x+n} \choose {x}}\right)}$ necessarily divides $x+j$.
Edit: Cleaning up the question. I think that much of my original assumptions are unnecessary.
Edit 2: @Mathlove points out a mistake in my reasoning. For $x=5, n=4, p=2$ for example, $v_p\left({{x+n}\choose{x}}\right) < t$
This is consistent with my conclusion even if it changes the details of my argument.
This question relates to step(6) in a previous question.