Show that $\prod_{1\le p\le n}p\le4^n$ using the fact that $\prod_{m+1\le p\le 2m}p\le {2m\choose m}$ where $p$ are prime.
I managed to show that $\prod_{m+1\le p\le 2m}p\le {2m\choose m}$, and I sort of know how this $4^n$ is achieved by similar processes, but I am having a problem making my way to those expressions exactly.
Attempt(which I hope you could relate to): Let us denote the set of prime numbers in $[m,n]$ as $\pi(m,n)$. Then: $\pi(1,n)=(\pi(\lfloor{n\over 2}\rfloor+1,2\lfloor{n\over 2}\rfloor)\cup\pi(2\lfloor{n\over 2}\rfloor+1,n))\cup(\pi(\lfloor{n\over 4}\rfloor+1,2\lfloor{n\over 2}\rfloor)\cup\pi(2\lfloor{n\over 4}\rfloor+1,\lfloor{n\over 2}\rfloor))\cup...\cup(\pi(\lfloor{n\over 2^i}\rfloor+1,2\lfloor{n\over 2^i}\rfloor)\cup\pi(2\lfloor{n\over 2^i}\rfloor+1,\lfloor{n\over 2^{i-1}}\rfloor))\cup...$.
Where $\pi(2\lfloor{n\over 2^i}\rfloor+1,\lfloor{n\over 2^{i-1}}\rfloor))$ is a singleton: for $i=1$ it is clear because if $a=2b+1$ then $2\lfloor{a\over 2}\rfloor+1=a=\lfloor{a\over 2^0=1}\rfloor$, and it keeps that way inductively.
First, let us agree that for a set $S$ with no prime numbers, $\prod_{p\in S}p=1$, so as to enable partitioning the intervals. Then $\prod_{1\le p\le n}p=\prod_{p\in\pi(\lfloor{n\over 2}\rfloor+1,2\lfloor{n\over 2}\rfloor)}p\prod_{p\in \pi(\lfloor{n\over 4}\rfloor+1,2\lfloor{n\over 4}\rfloor)}p...\le{2\lfloor {n\over 2}\rfloor\choose \lfloor {n\over 2}\rfloor}\cdot {2\lfloor {n\over 4}\rfloor\choose \lfloor {n\over 4}\rfloor}\cdot {2\lfloor {n\over 8}\rfloor\choose \lfloor {n\over 8}\rfloor}\cdot...$.
Here, I am kinda stuck, and not being fully sure of what I have done so far, I am going to need some guidance, thank you in advance.
