Let $x,n$ be integers with $x \ge n > 2$
Let $v_p(n)$ be the highest power of $p$ such that $p^{v_p(n)} \le n < p^{v_p(n)+1}$.
Assume that for a given prime $p$ such that $2 < p \le n$, none of the $n$ integers $x+1, x+2, \dots, x+n$ is divisible by $p^{v_p{n}+1}$
Does it then follow that $p^{v_p{n}+1} \nmid {{x+n} \choose n}$?
Here's an example of what I am talking about:
Let:
$x = 11$
$n = 10$
$p = 3$
$v_3(10) = 2$, and $27$ does not divide any integer in the set $\{ 11, 12, \dots, 21 \}$
Likewise, $27$ doesn't divide ${ {21} \choose {10}}$
To analyze this case, we only need to consider the following ratio: $\frac{3*3*9*3}{3*3*9}$. In this case, $9 \nmid {{21} \choose {10}}$.
I suspect that it is a simple argument to generalize this result. Am I wrong?
Edit 1: I've had some thoughts which, if true (I still need to prove them), would resolve my question. I believe all of these points are true.
I would greatly appreciate it if either someone can point out where I am wrong or help me to understand how to improve my argument so that it is less wordy and more mathematically precise.
Note: In all cases, I'm assuming that no integer in $\{1, 2, \dots, n\}$ is divisible by $p^{v_p(n)+1}$ and no integers in $\{x+1, x+2, \dots, x+n\}$ is divisible by $p^{v_p(n)+1}$
(1) If the number of integers divisible by $p$ in the set $\{ 1, 2, \dots n \}$ is the same as the number of integers divisible by $p$ in the set $\{x+1, x+2, \dots, x+n\}$, then the maximum power of $p$ that divides ${{x+n} \choose {n}}$ is $p^{v_p{n}-1}$.
(2) If the number of integers divisible by $p$ in $\{1, 2, \dots n\}$ is different than the number of integers divisible by $p$ in $\{x+1, x+2, \dots, x+n \}$, then the difference in number of integers divisible by $p$ is $1$.
(3) If the number of integers divisible by $p$ in $\{1, 2, \dots n\}$ is less than the number of integers divisible by $p$ in $\{x+1, x+2, \dots x+n\}$, then the maximum power of $p$ that divides ${{x+n} \choose {n}}$ is $p^{v_p{n}}$.