> [The Euler Product Formula] states that the sum of the zeta function is equal to the product of the reciprocal of one minus the reciprocal of primes to the power s. This astonishing connection laid the foundation for modern prime number theory...Here is the Euler formula for middle school.
We start with an infinite product of power series for all primes:
(1 + 2 + 2^2 + ... + 2^k + ...) *
(1 + 3 + 3^2 + ... + 3^k + ...) *
...
(1 + p + p^2 + ... + p^k + ...) *
...
Let's open the parentheses in an orderly manner without running into infinity too prematurely
(i.e. taking only a finite number of non-unit terms in each product):
1 + 2 + 3 + 2^2 + 5 + 2 * 3 + 7 + 2^3 + 3^2 + 2 * 5 + 11 + 2^2 * 3 + ...
Now, if we apply any completely multiplicative function f(ab) = f(a)f(b) to the terms we get
(1 + f(2) + f(2^2) + ... + f(2^k) + ...) *
(1 + f(3) + f(3^2) + ... + f(3^k) + ...) *
...
(1 + f(p) + f(p^2) + ... + f(p^k) + ...) *
... =
1 + f(2) + f(3) + f(4) + f(5) + f(6) + ...
When f(n) = n^(-s) we get the zeta function on the right and an infinite product of converging sums (s > 1) on the left.
The converging sums S = 1 + p^(-s) + p^(-2s) + ... are of course 1/(1 - p^(-s)) as S = 1 + S * p^(-s).
Voilà,
1 + 2^(-s) + 3^(-s) + ... = 1/(1 - 2^(-s)) * 1/(1 - 3^(-s)) * ... * 1/(1 - p^(-s)) ...
In case you think the Möbius formula is totally beyond the middle school curriculum ("primitive roots" and stuff):
SUM m(n) * n^(-s) = (1 - 2^(-s))(1 - 3^(-s))(1 - 5^(-s))...(1 - p^(-s))...
(just open the parentheses on the right, that's what middle-schoolers do).
And what happens if we multiply our power series S by 1 - p^(-s)? They cancel each other, so the whole product is 1.
The "astonishing connection" of the Euler formula turns out to be the fundamental theorem of arithmetic in disguise.