Can be justified this formula for $\zeta(2n+1)$

Question

Can be justified for integers $n\geq 1$ that

$$\zeta(2n+1)=\prod_{\text{p, prime}}\frac{1-\sigma(p^{2n})^{-1}}{1-p^{-2n}}?$$

Truly I don't know if I am wrong another time, when I use for an integer $m\geq 2$ that $$\zeta(m)=\prod_{\text{p, prime}}(1-p^{-m})^{-1}$$ to prove $$\frac{\zeta(m+1)}{\zeta(m)}=\prod_{\text{p, prime}}(1-\frac{1}{\sigma(p^m)}),$$ where $\sigma(l)$ is the sum of divisors function, and now using that Euler's product for Mobius function is $1/\zeta(m)$ we can conclude with the susbstitution $m=2n$.

Question. Please, can be justified or there is a mistake, that for integers $n\geq 1$ $$\zeta(2n+1)=\prod_{\text{p, prime}}\frac{1-\sigma(p^{2n})^{-1}}{1-p^{-2n}},$$ taking $\sigma(l)$ the sum of divisors function?

Answer 1

For any prime $p$, we have $$ \sigma(p^{2n}) = \frac{p^{2n+1}-1}{p-1} $$ and so $$ \frac{1-\sigma(p^{2n})^{-1}}{1-p^{-2n}} = \frac{1-\frac{p-1}{p^{2n+1}-1}}{1-p^{-2n}}=\frac{p^{2n+1}}{p^{2n+1}-1} = \left( 1 - \frac{1}{p^{2n+1}} \right)^{-1}. $$ Hence, $$\prod_{p, \text{prime}} \frac{1-\sigma(p^{2n})^{-1}}{1-p^{-2n}} = \prod_{p, \text{prime}} \left( 1 - \frac{1}{p^{2n+1}} \right)^{-1}=\zeta(2n+1). $$