We begin with the well-known Euler product form of the Riemann zeta function, an alternative to the summation definition. We can then expand it into a series:$$ \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p} \frac{1}{1-p^{-s}} = \prod_{p} 1+ \frac{1}{p^s-1} = \biggl(1+\frac{1}{p_{1}^s-1}\biggl)\biggl(1+\frac{1}{p_{2}^s-1}\biggl)\dots\biggl(1+\frac{1}{p_{n}^s-1}\biggl) $$At this point we can substitute $a_n$ with $p_{n}^s-1$, simplifying the notation:$$ p_{n}^s-1 = a_n \to \biggl(1+\frac{1}{a_1}\biggl)\biggl(1+\frac{1}{a_2}\biggl)\biggl(1+\frac{1}{a_3}\biggl)\biggl(1+\frac{1}{a_4}\biggl)\dots\biggl(1+\frac{1}{a_n}\biggl) = $$$$ = \biggl[1+\biggl(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+\dots\biggl)+\biggl(\frac{1}{a_1a_2}+\frac{1}{a_1a_3}+\frac{1}{a_2a_3}+\frac{1}{a_2a_4}+\frac{1}{a_3a_4}+\frac{1}{a_1a_4}+\dots\biggl)+ $$$$\biggl(\frac{1}{a_1a_2a_3}+\frac{1}{a_1a_2a_4}+\frac{1}{a_1a_3a_4}+\frac{1}{a_2a_3a_4}+\dots\biggl) + \biggl(\frac{1}{a_1a_2a_3a_4}\biggl)\biggl]\biggl(1+\frac{1}{a_n}\biggl) $$ \Writing each level of expansion as a sum allows us to use notation that expresses the increasing number of factors in each term. The first sum is simply:$$ \biggl(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+\dots\biggl) = \sum_{i=1}^{n} \frac{1}{a_i} $$This is a simple sum of inverses. However the following series will need a representation that takes into count the permutations of indexes, so we'll have, starting from one element, ($a_i$) up to three ($a_ia_ja_k)$:$$ \sum_{i=1}^{n} \frac{1}{a_i} \rightarrow \sum_{1\leq i<j\leq n} \frac{1}{a_ia_j} \rightarrow \sum_{1\leq i<j<k\leq n} \frac{1}{a_ia_ja_k} $$\justifying{This pattern continues, with each term represented as a sum of all products of $k$ distinct terms. For example, with $n=4$, taking terms where $1\leq i<j\leq 4$ yields six possible pairs. This happens because the number of permutations can be calculated using the binomial coefficient $ \binom{n}{k} $ where n is the number of terms chosen and k is the number of elements (for n=4 of k=2 the binomial coefficient is $\binom{4}{2} = 6$).}In general, the form of each term can be expressed as:$$ \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} \frac{1}{a_{i_1} a_{i_2} \cdots a_{i_k}}$$Thus, we can express the sum as:$$ 1+\sum_{k=1}^{n} \biggl(\sum_{1 \leq i_1 < \cdots < i_k \leq n} \frac{1}{a_{i_1} \cdots a_{i_k}}\biggl)+\prod_{n=1}^{\infty} \frac{1}{a_n}$$
The product term tends to zero as $n\to\infty$, shown as follows:$$ \lim\limits_{k\to\infty}\prod_{n=1}^{k} \frac{1}{p_{n}^s-1} = \lim\limits_{k\to\infty}\exp\biggl[\biggl(-\sum_{n=1}^{k} \ln\biggl(\frac{1}{p_{n}^s-1}\biggl)\biggl] = \lim\limits_{k\to\infty} e^{-k} = 0$$This follows because the sequence diverges as $k\to\infty$, leading the exponential term to tend to zero.With the product term being simplified, we are left with the following identity:$$ \zeta(s) = 1+\sum_{k=1}^{n} \sum_{1 \leq i_1 < \cdots < i_k \leq n} \frac{1}{a_{i_1} \cdots a_{i_k}}$$Subbing in $p^s_{n}-1 = a_n$ and applying the limit to infinity for accuracy:$$ \zeta(s) = 1+\lim\limits_{n\to\infty}\sum_{k=1}^{n} \sum_{1 \leq i_1 < \cdots < i_k \leq n} \frac{1}{(p^s_{i_1}-1) \cdots (p^s_{i_k}-1)}$$
Are there any mistakes or typos in this demonstration?