In Wikipedia's proof of Bertrand's Postulate, in the second lemma, it is concluded that:
$$R = R(p,{{2n}\choose{n}}) \le \log_p 2n$$
where $R(p,n)$ is the p-adic order of ${2n}\choose{n}$
Later in the main proof, the implication is:
$$\prod\limits_{p \le \sqrt{2n}}p^{R(p,{{2n}\choose{n}})} \le (2n)^{\pi(\sqrt{2n})}$$
It seems to me that there is a slightly stronger upper bound that is also implied.
$$\prod\limits_{p \le \sqrt{2n}}p^{R(p,{{2n}\choose{n}})} \le \frac{(2n)!}{(2n - \pi(\sqrt{2n}))!}$$
Am I wrong?
Here's my thinking:
(1) For each prime $p \le \sqrt{2n}$, there exists one unique number $p^{R(p,{{2n}\choose{n}})}$ that is less than or equal to $2n$.
(2) Since each $p^{R(p,{{2n}\choose{n}})} \le 2n$ and distinct, it follows that $\prod\limits_{p \le \sqrt{2n}} p^{R(p,{{2n}\choose{n}})}$ will be less than or equal to $\frac{(2n)!}{(2n - \pi(\sqrt{2n}))!}$
Edit: I made the changes pointed out by John Omielan.
Your slightly stronger upper bound, and your thinking regarding it, are both correct. Good work. Nonetheless, FWIW, here's an alternate way to explain your (2). Let $\sigma$ be a permutation of $p^{R(p,{{2n}\choose{n}})}$ for all primes $p \le \sqrt{2n}$, with the values being in strictly decreasing order. With $m = \pi(\sqrt{2n})$, we therefore have
$$2n \ge \sigma_{1} \gt \sigma_{2} \gt \ldots \gt \sigma_{m} \;\;\to\;\; \sigma_{i} \le 2n - (i - 1) \; \forall \; 1 \le i \le m$$
Thus, by the commutative property of multiplication, we then get
$$\prod_{p \le \sqrt{2n}}p^{R(p,{{2n}\choose{n}})} = \prod_{i=1}^{m}\sigma_{i} \le \\ \prod_{i=1}^{m}(2n - i + 1) = (2n)\cdots(2n-\pi(\sqrt{2n})+1) = \frac{(2n)!}{(2n - \pi(\sqrt{2n}))!}$$