Riemann Zeta circularity?

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In this post I show: $$\prod _{p\text{ prime}} \frac{p^s}{p^s-1} = \zeta(s).$$ Wolfram Alpha shows an alternate form for the primes: $$\frac{p_n{}^2}{p_n{}^2-1}=\frac{\left(\sum _{k=1}^{2^n} \left\lfloor \left(\frac{n}{\pi (k)+1}\right)^{1/n}\right\rfloor +1\right){}^2}{\left(\sum _{k=1}^{2^n} \left\lfloor \left(\frac{n}{\pi (k)+1}\right)^{1/n}\right\rfloor +1\right){}^2-1}, $$ and I note the prime counting function. I read somewhere that $\zeta(1)$ is equivalent to the Prime Number Theorem, so I have this question.

Would using primes to prove RH be circular? Or, is the fact that $s=1$ is a pole in the complex plane, mean there is no circularity?

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Both identities can be proved directly, without reference to the prime number theorem or the Riemann hypothesis or each other. So there's no circularity.