Question in Page 48 of Elementary Number theory by David Burton I am self studying Elementary number theory by David Burton's book and I could not think about how to prove this inequality . 
I am adding its image  highlighting the inequality I am not able to prove.
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Proof of Bonse Inequality is easy by using  2nd Principle of induction and Assuming Bertrand Postulate. 
But I am unable to prove the highlighted inequality assuming Bertrand postulate and usind 2nd principle of induction.

Can someone please tell how to prove it. 
 A: The formulation of Bertrand's postulate which is most useful here is that, for $n \ge 1$,
$$p_{n+1} \lt 2p_n \implies 2p_{n+1} \lt 4p_n \tag{1}\label{eq1A}$$
Increasing $n$ by $1$ and using \eqref{eq1A} gives
$$p_{n+2} \lt 2p_{n+1} \lt 4p_n = 2^2p_n \tag{2}\label{eq2A}$$
You can use induction (which I'll leave to you) to show that, for any $k \ge 1$, you have
$$p_{n+k} \lt 2^k p_{n} \implies p_{2n} \lt 2^n p_n \tag{3}\label{eq3A}$$
You have $p_2 p_3 p_4 = 3(5)(7) = 105 \gt 2^6 = 64$. As all other primes with higher indices are greater than $2$, you have for $n \ge 5$ that
$$\begin{equation}\begin{aligned}
p_2p_3 \cdots p_{n-1}p_n - 2 & \gt 2^6(p_5p_6 \cdots p_{n-1})p_n - 2 \\
& \gt 2^6\left(2^{n-5}\right)p_n - 2 \\
& = 2^{n+1}p_n - 2 \\
& \gt 2^{n}p_n \\
& \gt p_{2n}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
As this is for $n \ge 5$, you now just have to check the specific cases of $n = 3$, which gives $p_6 = 13$ and $p_2 p_3 - 2 = 3(5) - 2 = 13$, and $n = 4$, which gives $p_8 = 23$ and $p_2 p_3 p_4 - 2 = 3(5)(7) - 2 = 103$. Thus, this confirms
$$p_{2n} \le p_2 p_3 \cdots p_n - 2, \; n \ge 3 \tag{5}\label{eq5A}$$
is always true, with the equality only occurring for the $n = 3$ case.
