Is this formula correct $\sum_{n=2}^{\infty}\frac{1}{\pi^s(n)}\left\lfloor \frac{\phi(n)}{n-1}\right\rfloor=\zeta (s)$?

Euler's phi function $\phi(n)$ is the number of non-negative integers less than n that are relatively prime to n. The first few values of $\phi(n)$ for n=1,2,3,... are 1,1,2,2,4,...

$\pi(n)$ is prime counting function,

$\zeta(n)$ is Zeta function valid for Re(s)>1

We have arrived at a formula through the experimental of mathematics,

$$\sum_{n=2}^{\infty}\frac{1}{\pi^s(n)}\left\lfloor \frac{\phi(n)}{n-1}\right\rfloor=\zeta (s)$$

We need verification by proof that it is correct; can anyone offer us a proof?

Well, $\left\lfloor \frac{\phi(n)}{n-1}\right\rfloor$ is just an extraordinarily fancy way to say "1 when $n$ is prime and 0 otherwise". The rest is kinda trivial. Indeed, non-zero terms of this sum would correspond to the primes, and the values of $\pi(n)$ for them are just consecutive positive integers. $$\sum_{n=2}^{\infty}\frac{1}{\pi^s(n)}\left\lfloor \frac{\phi(n)}{n-1}\right\rfloor=\sum_{k=1}^{\infty}\frac{1}{\pi^s(\mathrm{prime}(k))}=\sum_{k=1}^{\infty}\frac{1}{k^s}=\zeta (s)$$