Csdrhrt

In the first line of Chapter Three of my graduate level text from which I am working from, the author declares the following divisibility relation to be the basis of Chebyshev theorem,
but has also been referred to as Bertrand's postulate:

$$S_0: n lt p_j lt 2n Rightarrow p_j quadBigg|quad {2nchoose n}$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(BCT0)}$$

Because I struggled with the first few exercises I attempted in my understanding of the authors justifications for steps taken, I decided that I must rigorously prove this introductory lemma before moving forward.
The first step I took, was an attempt at minimization of the the factors in the divisiblilty relation, instead of working with the factorials of the binomial coefficient in our original statement, obtaining a lemma that involves the lowest common multiples for the first $2n$ and $n$ natural numbers,which directly follows from our initial declaration, and so I found the proceding statement must also
be true if and only if the statement $S_0$ is true:

$$S_1: n lt p_j lt 2n Rightarrow p_j quadBigg|quad frac{operatorname{lcm}(1,2,3,...,2n)}{(operatorname{lcm}(1,2,3,...,n))^2} $$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(BCT1)}$$

<Important Additional Note Enclosed>

As one user has already commented, in the strictest sense that one states $a | b$ one requires $frac{b}{a} in mathbb Z$, in what I have stated here I at no point specified that $(operatorname{lcm}(1,2,3,...,n))^2 | operatorname{lcm}(1,2,3,...,2n)$ must be true.

When we speak of a divisibility relation $a | b$ in $a,bin mathbb Q$ as I have above, we no longer retain the assumptions as one makes in the conventional use of this type of lemma for regarding $a,bin mathbb Z$, rather than the requisite of $frac{b}{a} in mathbb Z$ we instead have the requisite of the denominator of $b$ remaining unchanged or reduced by the division by $a$, but never is there an increase in it's total number of factors.

<Important Additional Note Enclosed>

We then consider the unique prime factorization product for the lowest common multiple expressions in $S_1$ from the consideration of the arithmetic law and the exponents of the terms in these unique factorization products implied
by the identity for the lowest common multiple in terms of the second Chebyshev function:

$$psi ( x ) =sum _{j=1}^{pi ( x ) }Bigllfloor {frac {ln ( x ) }{ln ( p_{{j}} ) }}
Bigrrfloor ln ( p_{{j}} ) land ln(operatorname{lcm}(1,2,3,...,n))=psi (n) $$
$$Leftarrow Rightarrow$$
$$operatorname{lcm}(1,2,3,...,n)=prod_{j=1}^{pi(n)}p_j^{Bigllfloor frac{ln(n)}{ln(p_j)} Bigrrfloor} $$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(BCT2)}$$

This allows one to restate $S_1$ purely in terms of prime factorization products, in mutual consideration over which range our divisor product must be taken as per the inequality stated in both $S_0$ & $S_1$:

$$S_{3a}:prod_{j=pi(n)+1}^{pi(2n)-1}p_j quadBigg|quad frac{ prod^{pi(2n)}_{j=1}p_j^{bigllfloor frac{ln(2n)}{ln(p_j)} bigrrfloor}}{prod_{j=1}^{pi(n)}p_j^{2bigllfloor frac{ln(n)}{ln(p_j)} bigrrfloor}}$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(BCT3.0)}$$

Let the numerator of $frac{operatorname{lcm}(1,2,3,...,2n)}{(operatorname{lcm}(1,2,3,...,n))^2}$ be denoted as $mathcal N_n$.

We can always be assured that:

$$S_{3b}: p_{pi(2n)}=frac{mathcal N_n}{prod_{j=pi(n)+1}^{{pi(2n)-1}}p_j} $$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(BCT3.1)}$$

From $S_3$ we can easily then construct an assertion regarding a finite set that dictates that a necessary requisite for $S_1$ (hence also our original statment $S_0$) to be true, $S_1$ is true if and only if it is true that for any $n$ greater than $1$:

$$S_4: minBiggl({Bigl{Bigllfloor frac{ln(2n)}{p_j} Bigrrfloor-2Bigllfloor frac{ln(n)}{p_j} Bigrrfloor}Bigr}_{j=pi(n)+1..pi(2n)-1}Biggr)=1$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(BCT4)}$$

We finally note the following to be true for all $n$ greater than $1$:

$$S_5: {Bigl{Bigllfloor frac{ln(2n)}{p_j} Bigrrfloor-2Bigllfloor frac{ln(n)}{p_j} Bigrrfloor}Bigr}_{j=pi(n)+1..pi(2n)-1}={Bigl{1}Bigr} $$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(BCT5)}$$

$S_5$ assuring that $S_4$ is true for all $n$ greater than 1, hence also $S_1$ and our original statement $S_0$ must also be true for all $n$ greater than 1 and our proof is complete.