Proved by Sárkzözy [Sa85] for all sufficiently large $n$, and by Granville and Ramaré [GrRa96] for all $n\geq 5$.
More generally, if $f(n)$ is the largest integer such that, for some prime $p$, we have $p^{f(n)}$ dividing $\binom{2n}{n}$, then $f(n)$ should tend to infinity with $n$. Can one even disprove that $f(n)\gg \log n$?
Sander [Sa92] proved that $f(n)\to \infty$, and later [Sa95] quantified this to $f(n) \gg (\log n)^{1/10-o(1)}$. Sander [Sa95] also proved that $f(n)\ll \log n$ for all $n$, and that $f(n) \gg \log n$ for almost all $n$. The proofs of the latter two facts are very easy using Kummer's theorem -- indeed, this immediately implies that any $p$ divides $\binom{2n}{n}$ with multiplicity $\ll \log_p n$, and that the multiplicity with which $2$ divides $\binom{2n}{n}$ is equal to the number of $1$s in the binary expansion of $n$, which is $\gg \log n$ for almost all $n$.
Erdős and Kolesnik [ErKo99] improved the lower bound for all $n$ to \[f(n) \gg (\log n)^{1/4-o(1)}.\]
It remains open whether $f(n) \gg \log n$ for all $n$.
This is equivalent (via Kummer's theorem) to whether there are infinitely many $n$ which have only digits $0,1$ in base $3$, digits $0,1,2$ in base $5$, and digits $0,1,2,3$ in base $7$.
The sequence of such $n$ is A030979 in the OEIS.
A positive answer would imply that \[\sum_{p\leq n}1_{p\mid \binom{2n}{n}}\frac{1}{p}=(1-o(1))\log\log n,\] and Erdős, Graham, Ruzsa, and Straus say there is 'no doubt' this latter claim is true.
Weisenberg has provided four easy examples that show Erdős and Graham were too optimistic here: \[\binom{7}{3}=5\cdot 7,\] \[\binom{10}{4}= 2\cdot 3\cdot 5\cdot 7,\] \[\binom{14}{4} = 7\cdot 11\cdot 13,\] and \[\binom{15}{6}=5\cdot 7\cdot 11\cdot 13.\]
Pomerance [Po14] has shown that for any $k\geq 0$ there are infinitely many $n$ such that $n-k\mid\binom{2n}{n}$, although the set of such $n$ has upper density $<1/3$. Pomerance also shows that the set of $n$ such that \[\prod_{1\leq i\leq k}(n+i)\mid \binom{2n}{n}\] has density $1$.
The smallest $n$ for each $k$ are listed as A375077 on the OEIS.
Erdős [Er79d] writes it 'seems certain' that this holds for every $c>0$, with only a finite number of exceptions (depending on $c$). Standard heuristics on prime gaps suggest that the largest prime divisor of $\binom{n}{k}$ is in fact \[> \min(n-k+1, e^{c\sqrt{k}})\] for some constant $c>0$.
This is essentially equivalent to [961].
Let $f(n)$ be the smallest $k$ such that $u>n^2$. Give bounds for $f(n)$.
Mahler's theorem implies $f(n)\to \infty$ as $n\to \infty$, but is ineffective, and so gives no bounds on the growth of $f(n)$.
One can similarly ask for estimates on the smallest integer $f(n,k)$ such that if $m$ is the factor of $\binom{n}{k}$ containing all primes $\leq f(n,k)$ then $m > n^2$.
This was resolved by Bergman [Be11], who proved that for any $2\leq i<j\leq n/2$ \[\textrm{gcd}\left( \binom{n}{i},\binom{n}{j}\right) \gg n^{1/2}\frac{2^i}{i^{3/2}},\] where the implied constant is absolute.
Erdős and Szekeres further conjectured that $p\geq i$ can be improved to $p>i$ except in a few special cases. In particular this fails when $i=2$ and $n$ being some particular powers of $2$. They also found some counterexamples when $i=3$, but only one counterexample when $i\geq 4$: \[\textrm{gcd}\left(\binom{28}{5},\binom{28}{14}\right)=2^3\cdot 3^3\cdot 5.\]
It is known that $f(n)=n/P(n)$ when $n$ is the product of two primes. Another example is $n=30$.
For the second problem, it is easy to see that for any $n$ we have $f(n)\geq p(n)$, where $p(n)$ is the smallest prime dividing $n$, and hence there are infinitely many $n$ (those $=p^2)$ such that $f(n)\geq n^{1/2}$.
Both Erdős and Singmaster believed the answer to this question is no, and in fact that there exists an absolute upper bound on the number of solutions.
Matomäki, Radziwill, Shao, Tao, and Teräväinen [MRSTT22] have proved that there are always at most two solutions if we restrict $k$ to \[k\geq \exp((\log n)^{2/3+\epsilon}),\] assuming $a$ is sufficiently large depending on $\epsilon>0$.