I am reading Fundamentals of Number Theory by LeVeque, and got stuck with proving Theorem 6.11:
$$ \prod_{p \in \mathbb{P} \cap [2, x]} \left(1-\frac{1}{p} \right)<\frac{1}{\ln{x}}, $$ where $ x \ge 2 $ and $\mathbb{P}$ is the set of prime numbers.
In the book, it is first shown that $$ \prod_{p \in \mathbb{P} \cap [2, x]} \left(\frac{1}{1-1/p} \right)> \ln{x}. $$ To show this, it is concluded that
$$ \prod_{p \in \mathbb{P} \cap [2, x]} \left(\frac{1}{1-1/p} \right)>\sum_{k=1}^{x} \frac{1}{k}. \tag{$*$} $$ But this $(*)$ is precisely the part which I am unable to show/understand. I tried writing
$$ \prod_{p \in \mathbb{P} \cap [2, x]} \left(\frac{1}{1-1/p} \right) = \text{exp} \left(\sum_{p \in \mathbb{P} \cap [2, x]} \ln \left(\frac{1}{1-1/p} \right) \right), $$ but that got me nowhere.
So how can $(*)$ be concluded?
Just to expand on Gary's comment:
This is a classic proof idea relating two different forms of the Riemann zeta function. The idea is to expand $\frac{1}{1-1/p}$ as a geometric series. $$\prod\limits_{p\le m \text{ prime}} \frac{1}{1-1/p}= \prod\limits_{p\le m \text{ prime}} \sum\limits_{j\ge 0} \frac{1}{p^j} = \sum\limits_{j_2,j_3\ldots,j_q \ge 0} \frac{1}{2^{j_2}3^{j_3}\cdots k^{j_k}}$$ where the denominator on the right hand side has all primes less than or equal to $m$.
As the powers $j_2,\ldots,j_k$ run from $0$ upward, the denominator on the right hand side will pass through every number whose prime factors are all less than or equal to $m$. Meanwhile, in the sum $$\sum\limits_{n =1}^m \frac{1}{n},$$ the numbers in the denominator are all less than or equal to $m$, so their prime factors are as well, and therefore every term in this sum is contained in the previous sum as well.