Proving claim regarding upper bound of primorial 
Let $\pi(m,n)$ be the number of primes in the interval $[m,n]$.

*

*Show that $\displaystyle \prod\limits_{p\in \pi(m+1,2m)}p\le {2m\choose m}$.

*Use the previous item to show that $\displaystyle \prod\limits_{p\in \pi(1,n)}p\le 4^n$.

*Deduce from the previous item that $|\pi(1,n)|=\text{O}(n/\log n)$.


Part 1: let $p$ be a prime number such that $m+1<p<2m$. The expression $\displaystyle {2m\choose m}$ is an integer and we have $${2m\choose m}=\frac{(m+1)(m+2)\cdots(2m)}{m!}$$Now, $p$ divides the numerator and doesn't divide the denominator, thus it divides $\displaystyle {2m\choose m}$. This claim holds for any $p\in (m+1,2m)$, thus $$\left(\prod\limits_{p\in \pi(m+1,2m)}p\right)|{2m\choose m} \implies \prod\limits_{p\in \pi(m+1,2m)}p\le {2m\choose m}$$
Part 2: $\displaystyle [1,n]=\bigcup_{i=0}^{\log n-1}[2^i+1,2^{i+1}]$, hence $$\prod_{p\in\pi (1,n)}p=\prod_{i=0}^{\log n-1}\prod_{p\in\pi \left(2^{i+1},2^{i+1}\right)}p\underset{\text{By part 1}}{\le}\prod_{i=0}^{\log n-1}{2^{i+1}\choose 2^i}={2\choose 1}\cdot{{4\choose 2}}\cdots{n\choose \lfloor \frac{n}{2} \rfloor}$$Now, $\displaystyle {n\choose \lfloor \frac{n}{2} \rfloor}\le 2^n$, hence $$\prod_{p\in \pi(1,n)}p\le 2^2\cdot{2^4}\cdots 2^{2^{\log n}}=2^{2\left(1+2+4+\dots+\log n\right)}=2^{2n-2}=\frac{4^n}{4}$$
I have no idea how to deduce part 3. I thought about taking the logarithm of both sides of part 2, i.e $\displaystyle \log\left(\prod_{p\in \pi(1,n)}p\right)=\sum_{p\in\pi(1,n)}\log p\le n\log 4$ and $\displaystyle |\pi(1,n)|=\sum_{p\in\pi(1,n)}1=\sum_{p\in\pi(1,n)}\frac{\log p}{\log p}$, but it doesn't seem like I can get $\log n$ out of this sum.
Thanks!
 A: To get part 3 you have to use the first chevychev function. The first chebychev function is defined as:
$$\theta(x)=\sum_{p\le x}\ln(p)$$
where $p$ is a prime number.
Now :
$$\theta(x)=\ln(\prod_{p\le x} p)$$
As you have shown that:
$$\prod_{p\le x}p\le 4^x$$
which means:
$$\theta(x)\le 2x\ln(2)$$
Now to connect $\theta(x)$ to the prime counting function we use Abel summation formula which is:
$$\sum_{k=\lceil{y}\rceil}^{\lfloor{x}\rfloor}a_{k}f(k)=f(x)\sum_{k=1}^{\lfloor{x}\rfloor}a_{k}-f(y)\sum_{k=1}^{\lfloor{y}\rfloor}a_{k}-\int_{y}^{x}f'(t)\sum_{k=1}^{\lfloor{t}\rfloor}a_{k}dt$$
So if we set $f(t)=\frac{1}{\ln(t)}$,$y=1.5$  &
$$ a_{k}=
    \begin{cases}
        0 & k\neq p \\
        \ln(p) & k= p \\
    \end{cases}
$$
Then we get:
$$\sum_{p\le x}1=\frac{1}{\ln(x)}\sum_{p\le x}\ln(p)+\int_{1.5}^{x}\frac{1}{t\ln^{2}(t)}\sum_{p\le t}\ln(k)dt$$
Thus:
$$\pi(x)=\frac{\theta(x)}{\ln(x)}+\int_{1.5}^{x}\frac{\theta(t)}{t\ln^{2}(t)}dt$$
Multiplying both sides by $\frac{\ln(x)}{x}$ we get:
$$\frac{\pi(x)\ln(x)}{x}=\frac{\theta(x)}{x}+\frac{\ln(x)}{x}\int_{1.5}^{x}\frac{\theta(t)}{t\ln^{2}(t)}dt$$
As $$\theta(x)\le 2x\ln(2)$$
$$\frac{\pi(x)\ln(x)}{x} \le 2\ln(2)+\frac{\ln(x)}{x}\int_{1.5}^{x}\frac{\theta(t)}{t\ln^{2}(t)}dt$$
As $\theta(x)=0 \forall x \in (0,2)$
$$\int_{1.5}^{x}\frac{\theta(t)}{t\ln^{2}(t)}dt=\int_{2}^{x}\frac{\theta(t)}{t\ln^{2}(t)}dt \le \int_{2}^{x}\frac{2t\ln(2)}{t\ln^{2}(t)}dt$$
$$\int_{2}^{x}\frac{2t\ln(2)}{t\ln^{2}(t)}dt=\int_{2}^{\sqrt{x}}\frac{2\ln(2)}{\ln^{2}(t)}dt+\int_{\sqrt{x}}^{x}\frac{2\ln(2)}{\ln^{2}(t)}dt$$
As $\frac{1}{\ln^{2}(t)}$ decreases monotonically:
$$\int_{2}^{\sqrt{x}}\frac{2\ln(2)}{\ln^{2}(t)}dt+\int_{\sqrt{x}}^{x}\frac{2\ln(2)}{\ln^{2}(t)}dt \le \int_{2}^{\sqrt{x}}\frac{2\ln(2)}{\ln^{2}(2)}dt+\int_{\sqrt{x}}^{x}\frac{2\ln(2)}{\ln^{2}(\sqrt{x})}dt =\frac{2(\sqrt{x}-2)\ln(2)}{\ln^{2}(2)}+\frac{2\ln(2)(x-\sqrt{x})}{\ln^{2}(\sqrt{x})}$$
Thus:
$$\frac{\pi(x)\ln(x)}{x} \le 2\ln(2)+\frac{2\ln(x)(\sqrt{x}-2)}{x\ln(2)}+\frac{2\ln(x)\ln(2)(x-\sqrt{x})}{x\ln^{2}(\sqrt{x})}$$
$$\frac{2\ln(x)(\sqrt{x}-2)}{x\ln(2)}+\frac{2\ln(x)\ln(2)(x-\sqrt{x})}{x\ln^{2}(\sqrt{x})}$$ converges to 0 as $x \rightarrow \infty$. It has no poles for $x>2$ which tells us that it has an upper bound call it $M$
Therefore:
$$\frac{\pi(x)\ln(x)}{x} \le 2\ln(2)+M$$
which means $\forall x>2$:
$$ |\pi{(x)}| \le \frac{(2\ln(2)+M)x}{\ln(x)}$$
This is the definition of big O-notation therefore we can conclude that:
$$\pi(x)=O{(\frac{x}{\ln(x)})}$$
