Let $c(n)=\frac{1}{n+1}\binom{2n}{n}$ be the Catalan number. It seems that a product $\prod_{n\in I} c(n)$, where $I\subset\mathbb N_{>1}$, is never a Catalan number. Is this a (known) fact?

1It should not be surprising, as Catalan numbers are divisible by almost all primes between n and 2n. Do you need a proof that a product does not produce such a number? Gerhard "Proudly, Proactively Proposing Productive Proposition" Paseman, 2018.12.05. – Gerhard Paseman Dec 6 at 6:43
It should be possible here to mimic the same argument that Erdos uses in his paper "On some divisibility properties of $\binom{2n}{n}$".
Suppose that $c(n)=c(a_1)c(a_2)\cdots c(a_k)$ and $n$ is large enough (not sure what constant is exactly needed here, but checking all $n\le 25$ should be enough). Since we can always find a prime $n+1<p<2n$, it must divide $c(n)$. Therefore we must have $$pc(a_i)$$ for some $i$, which in turn implies $a_i> n/2$. From here we want to show that it is impossible to have $\frac{c(n)}{c(m)}\in \mathbb Z$, with $\max(25,m)< n<2m$ giving us our contradiction.
To show this last part we can look at the fraction $$\frac{c(m+k)}{c(m)}=\frac{\frac{1}{(m+1)(m+k+1)}\prod_{i=1}^{2k} (2m+i)}{\prod_{i=2}^k (m+i)^2}$$ for integers $m>k>0$ with $m+k>25$.
In case $m<\frac{5}{6}(m+k)$ we have a prime $p$ that appears twice in the denominator and once in the numerator. In the opposite case, $m\geq 5k$ is enough to guarantee that the largest prime factor in the denominator is $\geq 2k$. If this largest prime appears with multiplicity $2\alpha$ in the denominator, it only appears with multiplicity $\alpha$ in the numerator. So either way the ratio can not be an integer.
Of course, I would consult with an actual number theorist about what the actual bounds should be, or if there can be more simplifications.

1Great! Experimentally, $n>5$ should be enough. (trivial typo: I think we are happy if $c(n)/c(m) \not\in\mathbb{Z}$ if $\max(25,m) < n < 2m$) – Martin Rubey Dec 6 at 9:07