<s>[INST] Show that the following identity is formally correct $$\sum_{k=0}^{\infty}\frac{1}{2^{2k}}\sum_{k=0}^{\infty}\frac{1}{3^{2k}}\sum_{k=0}^{\infty}\frac{1}{5^{2k}}\cdots=\sum_{n=1}^{\infty}\frac{1}{n^2}$$
Ignoring problems connected with convergence and rearrangement of terms, and the denominators on the left are the even powers of the primes.
 [/INST] Suppose you have an arbitrary finite set of primes, e.g. $\{2,3,5\}$.  Look at
$$
\left(1 + \frac{1}{2^2} + \frac{1}{2^4} + \frac{1}{2^6} +\cdots \right)\left(1 + \frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} +\cdots \right)\left(1 + \frac{1}{5^2} + \frac{1}{5^4} + \frac{1}{5^6} +\cdots \right).
$$
This expands to
$$
\sum_{a,b,c \ge 0} \frac{1}{2^{2a}} \frac{1}{3^{2b}} \frac{1}{5^{2c}} = \sum_{a,b,c \ge 0} \left(\frac{1}{2^a} \frac{1}{3^b} \frac{1}{5^c}\right)^2.
$$
The fraction runs throught the list of reciprocals of all possible products of $2$s, $3$s, and $5$s.
Now instead of just those primes, $2$, $3$, and $5$, use the list of all primes.  The set of all possible products of not-necessarily distinct primes is just the set of all positive integers.
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