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.

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.
