I have been thinking about the Sylvester and Schur theorem and wondering if there is an alternative argument.
Let $v_p(n)$ be the highest power of $p$ that divides $n$.
It occurs to me that:
$$v_p\left({{x+n}\choose{n}}\right) \le \max\left(v_p(x+1), v_p(x+2), \dots, v_p(x+n)\right)$$
Note: I have detailed my reasoning here.
If this assumption is correct, would the following be a valid argument for establishing that if $n \ge 8, x \ge \frac{n}{2}$, then a prime $p > n$ divides ${{x+n}\choose{n}}$:
(1) Let $T$ be the set $\{x + 1, x + 2, \dots, x+n\}$ so that:
$${{x+n}\choose{n}} = \frac{\prod\limits_{t \in T}t}{n!}$$
(2) Assume that no primes greater than $n$ divide ${{x+n}\choose{n}}$.
(3) Let $U = \prod\limits_{p \le n}p^{v_p\left({{x+n}\choose{n}}\right)}$ so that:
$${{x+n}\choose{n}} = U$$
(4) If my assumption is correct, there exists a set $S$ such that $S \subset T$ and $|S| = \pi(n)$ and there exists $U'$ such that $\prod\limits_{s \in S}s = UU'$, and $V = \dfrac{\prod\limits_{t \in T}t}{\prod\limits_{s \in S}s}$ so that:
$${{x+n}\choose{n}} = U = \frac{UU'V}{n!}$$
(5) As $x$ increases and $n$ stays the same, if there are no primes greater than $n$, then $U$ will change but $U'V$ will stay constant. But this is clearly impossible since $V$ does not stay constant and $U' \ge 1$.
(6) Let $w = n - \pi(n)$. At all times, the values of $V$ is constrained by:
$$\frac{(x+w)!}{x!} \le V \le \frac{(x+n)!}{(x+n-w)!}$$
(7) At $x = 0$, for example:
$$w! \le V \le \frac{n!}{(n-w)!}$$
(7) As $x$ increases, this inequality changes, and eventually the minimum bound of the higher $x$ is greater than the maximum bound of the lower $x$. At this point, the minimum value of $V$ has increased independently of $U$ and it follows that:
$${{x+n}\choose{n}} > U$$.
(8) The first point where this definitely occurs for $n\ge 8$ is when $x > \frac{n}{2}$:
It is well known for $n\ge 8$, $\pi(n) \le \frac{n}{2}$ so that:
$\frac{n}{2} + w > \pi(n) + w = \pi(n) + n - \pi(n) = n$
