On variations of Erdős squarefree conjecture: a conjecture involving the prime-counting function I'm inspired in the so-called Erdős squarefree conjecture, this section from Wikipedia, to state in this post a similar variation involving the prime-counting function, I think that it is more difficult than the exercise in my recent previous post.
In this post we denote the prime-counting function as $\pi(x)$.
We know the multiplicative formula for factorials, this section of Wikipedia.
Computational fact. In the segment of positive integers $1\leq n\leq 20000$, the integer $n=1198$ is the last integer for which $$\binom{2n}{\pi(n)}$$
is a square-free integer (has no repeated prime factors). 
The sequence of binomial coefficients $$\binom{2n}{\pi(n)}$$ having some repeated prime factors (when $n\geq 1$ runs over integers) starts as $$4,28,120,220,1820\ldots$$
From previous computational evidence (I know that isn't the best) I state the following conjecture.
Conjecture. The binomial coefficient $$\binom{2n}{\pi(n)}$$ is never squarefree for $n>1198$.

Question. Can you prove it or provide us a counterexample? If do you think that there exists a positive integer $n_0$ such that the conjecture holds for all $n>n_0$ tell us what work can be done about it. Many thanks.

Optionally, since I don't know if my proposal of variation of  Erdős squarefree conjecture is interesting/artificious, if you want to add some different conjecture or a critic, feel free to add a comment.
 A: This 1996 paper by A. Granville and O. Ramaré ("Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients", Mathematika 43 (1996), 73-107) contains the following

Theorem. There exists a constant $\tau_1>0$ such that if $n$ is sufficiently large and ${{n}\choose{k}}$ is squarefree then $k$ or $n-k$ is $<\exp\left(\tau_1(\log n)^{2/3}(\log\log n)^{1/3}\right)$. 

In other words, for large enough $n$, squarefree binomials coefficients aren't found for $k$ farther than
$$
\exp\left(\tau_1\log n\left(\frac{\log \log n}{\log n}\right)^{1/3}\right)=n^{\tau_1\left(\frac{\log \log n}{\log n}\right)^{1/3}}
$$
away from $0$ or $n$, which is $O(n^\varepsilon)$ for any positive $\varepsilon$.  On the other hand, the prime-counting function is $\pi(n) \sim n/\log n$, which is $\Omega(n^{1-\varepsilon})$ for any positive $\varepsilon$.  So your conjecture certainly holds for $n$ greater than some $n_0$... and it would also hold if $\pi(n)$ were replaced by any sequence growing fast enough with $n$.  
