Could you please have a look at my solution of the following exercise?
Show that for $s>1$ holds $$\sum_{n = 1}^{\infty} \frac{1}{n^s} = \prod_{k=1}^{\infty}\frac{1}{1-p_k^{-s}}$$
where $(p_k)_k$ is the increasing sequence of primes.
I came up with the following solution:
At first we note that
$$\prod_{k=1}^{\infty}\frac{1}{1-p_k^{-s}} = \prod_{k=1}^{\infty} \frac{p_k^s}{p_k^s-1} = \prod_{k=1}^{\infty} \biggl( \sum_{j=0}^{\infty} \frac{1}{p_k^{s \cdot j}}\biggr)$$
Now we show both $\ge$ and $\le$:
$\ge:$
$$\prod_{k=1}^{N}\frac{1}{1-p_k^{-s}} = \prod_{k=1}^{N} \frac{p_k^s}{p_k^s-1} $$
To make things easier we take care that the exponents are natural numbers
$$\le \prod_{k=1}^{N} \frac{p_k^{\lceil s \rceil}}{p_k^{\lfloor s \rfloor} -1}$$
For fitting natural numbers $A,B$ we may write
$$ = \frac{A}{B} = \sum_{n=1}^{A} \frac{1}{B} $$
And since $A \ge B$
$$\le \sum_{n=1}^{A} \frac{1}{n}$$
$\le:$
We make the observation that in the result of the multiplication
$$\biggl( \sum_{i=0}^{a} \frac{1}{p_1^i} \biggr) \cdot \biggl( \sum_{j=0}^{b} \frac{1}{p_2^j}\biggr) = \sum_{i=0}^{a} \sum_{j=0}^{b} \frac{1}{p_1^i \cdot p_2^j} $$
each combination of powers of $p_1$ and $p_2$ up to $p_1^a$ and $p_2^b$ appears exactly once in the denominators on the right side above. By generalising this and remembering about the fundametal theorem of arithmetic we obtain:
$$\sum_{n = 1}^{N} \frac{1}{n^s} \le \prod_{k=1}^{A} \biggl( \sum_{j=0}^{B_k} \frac{1}{p_k^{s \cdot j}}\biggr) $$
for some natural numbers $A$ and $B$ and further:
$$\le \prod_{k=1}^{A} \biggl( \sum_{j=0}^{\infty} \frac{1}{p_k^{s \cdot j}}\biggr) = \prod_{k=1}^{A} \frac{p_k^s}{p_k^s-1} \le \prod_{k=1}^{\infty} \frac{p_k^s}{p_k^s-1}$$
Is this correct and if yes, could you tell me where exactly we need that $s>1$; I do not see why this is required.