Are all denominators of $\zeta(2n)$ abundant, for $n\geq2$?

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Consider the Riemann zeta function $\zeta(n)$ defined for positive integers $n$ as usual by:

$$ \zeta (n)=\sum _{k=1}^{\infty }{\frac {1}{k^{n}}}. $$

For positive even numbers we have the identity:

$$ {\displaystyle A_{n}\zeta (2n)=B_{n}\pi ^{2n}\,\!}, $$

where the sequence $A_n$ ($A002432$ in the OEIS) represents the denominators in $\zeta(2s)$. Then I propose the following conjecture:

Conjecture 1. For all $n\geq2$, $A_n$ is abundant.

Is this known? Is there an explanation for this? I have verified the conjecture for $2\leq n \leq 17$.