Landau's proof that $\zeta(s)=\prod_{p} \frac{1}{1-p^{-s}}$

Question

In the Handbuch, Landau proves that for all $s>1$ the following equality holds $$\zeta(s)=\prod_{p} \frac{1}{1-p^{-s}}.$$ I'm having trouble with the following part of his proof:

Landau says that for all $s>1$ we have $$\prod_{p\leq x} \frac{1}{1-p^{-s}}=\prod_{p\leq x}\big(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+ ...\big)=\sum_{n=1}^{\infty}'\frac{1}{n^s}$$where in the last sum $n$ runs through all numbers whose prime factors are all $\leq x$.

Can someone please shed light on the last equality? I don't see how that product is equal to this special sum.

Thank you.

Answer 1

This is due to the Fundamental Theorem of Arithmetic that states, to quote Wikipedia, that "every integer greater than 1 either is prime itself or is the product of prime numbers, and that, although the order of the primes in the second case is arbitrary, the primes themselves are not."

Basically, having multiplied out the multiplicand of the product you obtain the reciprocal powers of the integers up to a value depending on $x$. Letting $x\to\infty$ gives you all possible reciprocal powers of the integers.

See also here for more information.

Answer 2

Try expanding the product with $x=3$ or $x=5$ to get a hands-on feel for why.

Answer 3

$\dfrac{1}{1-p^{-s}} = \sum_{k=0}^{+\infty}(\dfrac{1}{p^s})^k$.

Multiply all these $\dfrac{1}{1-p^{-s}}$ together gives all the possible terms in form of $\prod_{i=1}^n(\dfrac{1}{p_i^s})^{k_i}= \dfrac{1}{(\prod_{i=1}^n p_i^{k_i})^s}$, where $p_i$ are prime numbers less than $x$