Largest power of $p$ that divides $2n$ choose $n$?

I'm working through questions that I'm not familiar with and would like some help on the following:

If $$π(x)$$ counts the number of primes less than or equal to $$x$$, how many primes are there greater than $$n$$ but less than or equal to $$2n$$ ?

And if prime $$p$$ satisfies $$n < p ≤ 2n$$, what is the largest power of $$p$$ that divides $$2n$$ choose $$n$$ ?

Thanks!

As for the first question, the number of primes less than or equal to $$2n$$ is $$\pi(2n)$$ and the number of primes less than $$n$$ is $$\pi(n)$$; thus, the number of primes less than or equal to $$2n$$ but not less than or equal to $$n$$ is $$\pi(2n)-\pi(n)$$.
Secondly, recall that $${2n\choose n}=\frac{(2n)!}{n!^2}$$. Since $$p>n$$, we know $$p$$ cannot divide $$n!^2$$. Since there's only one multiple of $$p$$ between $$n$$ and $$2n$$ (because $$n, we know $$2n<2p$$), we know $$p$$ divides $$(2n)!$$ only once. Thus, $$p$$ divides $${2n\choose n}$$ exactly once.