The following proof is from the 19th page of
Everest, Graham; Ward, Thomas, An introduction to number theory, Graduate Texts in Mathematics 232. London: Springer (ISBN 1-85233-917-9/hbk). x, 294 p. (2005). ZBL1089.11001.
In fact, I think this proof is not finished. For the red line, only $k(p)\ge 2$ is disproved. For the case $k(p)=1$, there is not any talk. How to understand it ? Thanks.
$\textbf{Theorem 1.9.}\ [\text{B}{\scriptstyle{\text{ERTRAND'S}}} \text{ P}\scriptstyle{\text{OSTULATE}}]\ $ If $n\geqslant1$, then there is at least one prime $p$ with the property that $$n<p\leqslant2n.\tag{1.13}$$ $\text{P}\scriptstyle{\text{ROOF}}$. For any real number $x$, let $\lfloor x\rfloor$ denote the integer part of $x$. Thus $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$. Let $p$ be any prime. Then $$\left\lfloor\dfrac np\right\rfloor+\left\lfloor\dfrac n{p^2}\right\rfloor+\left\lfloor\dfrac n{p^3}\right\rfloor+\cdots$$ is the largest power of $p$ dividing $n!$ (see Exercise $8.7(a)$ on p. $162$). Fix $n\geqslant 1$ and let $$N=\prod_{p\leqslant2n}p^{k(p)}$$ be the prime decomposition of $N=(2n)!/(n!)^2$. The number of times that a given prime $p$ divides $N$ is the difference between the number of times it divides $(2n)!$ and $(n!)^2$, so $$k(p)=\sum_{m=1}^\infty\left(\left\lfloor\dfrac{2n}{p^m}\right\rfloor-2\left\lfloor\dfrac n{p^m}\right\rfloor\right),\tag{1.14}$$ and each of the terms in the sum is either $0$ or $1$, depending on whether $\left\lfloor\frac{2n}{p^m}\right\rfloor$ is odd or even. If $p^m>2n$ the term is certainly $0$, so $$k(p)\leqslant\left\lfloor\dfrac{\log2n}{\log p}\right\rfloor.\tag{1.15}$$
Best Answer
The idea is to notice that
$$\log(N) \leq \sum_{p|N}{\log(p)} + \sum_{k(p) \geq 2}{k(p)\log(p)}.$$
The first term is dealt with by $(1.16)$, the second one by the last estimate of the image before last.