Euler's famous expression for $\frac{\pi^2}6=\sum\frac1{n^2}$ is very elegant. I have also seen a second expression $$\frac{\pi^2}6=\prod_p\frac{p^2}{p^2-1}$$ that seems attributed to Euler too (the product is taken over all primes $p$). It is said that these two expressions are effectively equivalent.
I do note that the Wallis product seems to use Euler's infinite product for the sine function to obtain a value for $\frac\pi2$ and I suspect a similar approach could be used to obtain my second expression, though I cannot see how at present. Can anybody explain (or provide a link to an explanation of) how my product expression is derived and if it can be attributed to Euler (or obtained from his work)?
$$ \begin{align} \prod_{p\in\mathbb{P}}\frac{p^\alpha}{p^\alpha-1} &=\prod_{p\in\mathbb{P}}\frac1{1-\frac1{p^\alpha}}\\ &=\prod_{p\in\mathbb{P}}\sum_{k=0}^\infty\frac1{p^{k\alpha}}\tag1 \end{align} $$ Applying the Fundamental Theorem of Arithmetic, for any $n\in\mathbb{N}$, there is a unique product $$ n^\alpha=\prod_{j=1}^\infty p_j^{k_j\alpha}\tag2 $$ where only finitely many $k_j\ne0$. Thus, there is exactly one product of terms in $(1)$ corresponding to $n$. For example, $$ \color{#CCC}{\cdots+\frac1{11^\alpha}+}\frac1{12^\alpha}\color{#CCC}{+\frac1{13^\alpha}+\cdots} $$ corresponds to the term in the product $$ \color{#CCC}{\left(1+2^\alpha+\color{#000}{2^{2\alpha}}+\cdots\right)\left(1+\color{#000}{3^\alpha}+3^{2\alpha}+\cdots\right)\left(\color{#000}{1}+5^\alpha+5^{2\alpha}+\cdots\right)\cdots} $$ Therefore, adding up all the products of terms in $(1)$, we get $$ \prod_{p\in\mathbb{P}}\frac{p^\alpha}{p^\alpha-1} =\sum_{n=1}^\infty\frac1{n^\alpha}\tag3 $$