Proof of Infinitude of Primes by Euler's Product Formula is Circular? Many guides will refer to Euler's product formula as simple way to prove that the number of primes is infinite.
$$\sum_n\frac{1}{n} = \prod_p \frac{1}{1-\frac{1}{p}}$$
The argument is that if the primes were finite, the product on the right hand side is finite, noting that $1-\frac{1}{p}$ is never zero.
However, the product formula itself is constructed by application of the fundamental theorem of arithmetic to an infinite series with terms involving only primes.
Does this mean such proofs are a circular argument - because they use a product formula whose construction depends entirely on the infinitude of primes?
 A: As others already noted, it's non-circular because the FTA doesn't assume infinitely many primes. In fact, the original proof there are infinitely many primes uses a fragment of the FTA. We multiply together finitely many primes and add $1$, and to continue the argument we need to know the result will have some prime factor. We prove this helpful result early in the proof of the FTA.
A: To prove the equality you need that every natural number is uniquely represented as a product of primes. It does not need the fact that the set of primes is infinite. In fact to prove that the set of primes is infinite, you do not need Euler equality. You only need inequality $LHS \le RHS$ which follows from the fact that every number is a product of primes (uniqueness is not needed). If you compare that proof with Euclid's original proof, it is not clear which one uses less prior infomation about primes.
A: No, the proof is not circular. If we assume there are a finite number of primes $p_1,\ldots,p_k$, then we would assume that any $n\in\mathbb{N}$ would be able to be factorized as $p_1^{\alpha_1}\ldots p_k^{\alpha_k}$. (The proof of the Fundamental Theorem of Arithmetic does not require that there be an infinite number of primes.) This would lead to the Euler product formula, which we would then use to provide the contradiction, once we have shown that $\sum_{n\in\mathbb{N}} \frac1n$ is infinite.
