Inequality query in the book of G. H. Hardy 'An Introduction to the Theory of Numbers' Book: An Introduction to the Theory of Numbers - G. H. Hardy and Chapter XXII

The last inequality is the one I don't understand
Can you help give me a hint?
 A: From the first inequality we have $\rho\log\log n <\log n$. So $0\le\rho<(\log n)/\log\log n.$ And we have $ 0<1-1/\log n<1.$  So $$(1-1/\log n)^{\rho} >(1-1/\log n)^{(\log n)/\log\log n}.$$ The other term $\prod_{p\le\log n}(1-1/p)$ in the last line is exactly the same thing as the term $\prod_{i=1}^{r-\rho}(1-1/p_i)$ in the line above it.
A: Hardy & Wright's book was published many decades ago, and I believe the confusion can be addressed if we formulate this problem in terms of a new set of notations.
$$
{\varphi(n)\over n}=\prod_{p|n}\left(1-\frac1p\right)
$$
We may introduce a parameter $t>0$ to partition the product into two parts:
\begin{aligned}
{\varphi(n)\over n}
&=\prod_{\substack{p|n\\p\le t}}\left(1-\frac1p\right)\prod_{\substack{p|n\\p>t}}\left(1-\frac1p\right) \\
&>\left(1-\frac1t\right)^{\omega(n,t)}\prod_{p\le t}\left(1-\frac1p\right)
\end{aligned}
where $\omega(n,t)=\#\{p|n:p>t\}$. By the properties of factorizations, we know trivially that $t^{\omega(n,t)}<n$. This indicates that $\omega(n,t)<\log n/\log t$. Plugging this into the above expression, we conclude
$$
{\varphi(n)\over n}>\left(1-\frac1t\right)^{\log n/\log t}\prod_{p\le t}\left(1-\frac1p\right)
$$
Finally, setting $t=\log n$ gives the inequality in Hardy & Wright's book. By the way, happy halloween!
