Is this $\zeta(p)=\sum_{k\geq1 ,p\in \mathbb{P}}\frac{1}{{\sigma}_k(p)-1}$ true with sigma is power of sum divisor function?

Question

let ${\sigma}_k(n) =\sum_{d|n} d^k$ is a sum of divisor function , And let ${\sigma}_k(n)$ be the iterating divisor function, We have for every prime $p$ and for every integer $k\geq 1$ : $${\sigma}_k(p)=p^k+1 \tag1$$ , Now form $(1)$ we have :$$\sum_{k\geq1 ,p\in \mathbb{P}}p^{-k}= \sum_{k\geq1 ,p\in \mathbb{P}}\frac{1}{{\sigma}_k(p)-1}\tag2$$, then : $$\zeta(p)=\sum_{k\geq1 ,p\in \mathbb{P}}\frac{1}{{\sigma}_k(p)-1}\tag3$$ , Now my problem if i want to calculate for example $\zeta(2)$ I will get $1$ by $(3)$ ? then why that's not true however i have used correct formula ?