On an estimation of binomial coefficient On page 14 of the book 'Proofs from THE BOOK', there is an estimation presented as: 
$$\binom{2n}{n}\le \prod_{p\le \sqrt{2n}}\ 2n. \prod_{\sqrt{2n}<p\le \frac{2}{3}n}\ p. \prod_{n<p\le 2n}\ p, $$
with $n$ is a positive integer and $p$ is a prime number. 
I don't know how to obtain this result, and I think the factor $2n$ in the RHS must be $p$.
My proposed question is: How can the result be verified or proved?
 A: This results from bounding the number of possible occurrences of each prime in the numerator and denominator. For every integer $k$, the number of factors in the numerator divisible by $k$ can exceed the number of such factors in the denominator by at most $1$, since $\lfloor2n/k\rfloor-2\lfloor n/k\rfloor\le1$.
The primes from $\sqrt{2n}$ to $2n$ occur at most once in each factor, so there can be at most one factor $p$ in excess in the numerator. For $\frac23n\lt p\le n$, and thus $1\le\frac np\lt\frac32$, we have $\lfloor2n/p\rfloor-2\lfloor n/p\rfloor=0$, so these primes don't occur in the result at all. (MathWorld  says that the central binomial coefficient is divisble by $p$ iff the base-$p$ representation of $n$ contains no digits greater than $p/2$, a generalization of the fact used here.) 
The primes up to $\sqrt{2n}$ can occur more than once, but for each $k$ there can be at most one excess factor divisible by $p^k$, so the total excess of factors of $p$ is bounded by the maximal number of factors of $p$ occurring in any of the factors, and this is in turn bounded by the maximal factor, $2n$.
