Is $\sum\limits_{k=0}^{\infty}\frac{1}{(mk)!^{n+1}}$ irrational?

Question

I was using Wolfram Alpha for things, and I came across $I_{0}(2)$. For fun I tried asking Wolfram Alpha if the number was irrational, but said it's unknown. I believe this is an error, as its irrationality should already be confirmed, as by generalizing the Fourier's proof of the irrationality of $e$, you can prove that a whole class of numbers (That Wolfram Alpha considers unknown) is irrational, specifically

$A(m,n) = \sum\limits_{k=0}^{\infty}\frac{1}{(mk)!^{n+1}}$

is irrational. In particular,

$A(1,0) = e$

$A(1,1) = I_{0}(2)$ (Irrationality unknown for Wolfram Alpha)

$A(3,0) = \frac{1}{3}\left(e + \frac{2\cos\left(\frac{\sqrt 3}{2}\right)}{\sqrt e} \right)$ (Irrationality unknown for Wolfram Alpha)

$A(4,0) = \frac{1}{2}\left(\cos(1)+\cosh(1)\right)$ (Irrationality unknown for Wolfram Alpha)

For completeness, I'll add the generalized version of Fourier's proof:

If $A(m, n)$ is a rational number, then there exist $a_{m,n},b_{m,n} \in \mathbb{N}^+$ such that $A(m, n) = \frac{a_{m,n}}{b_{m,n}}$.

Let's define the number

$x_{m, n} = (mb_{m,n})!^{n+1}\left(A(m, n) - \sum\limits_{k=0}^{b_{m,n}}\frac{1}{(mk)!^{n+1}}\right)$

Under the previous assumption, we have

$x_{m, n} = (mb_{m,n})!^{n+1}\left(\frac{a_{m,n}}{b_{m,n}} - \sum\limits_{k=0}^{b_{m,n}}\frac{1}{(mk)!^{n+1}}\right) = \frac{a_{m,n}(mb_{m,n})!^{n+1}}{b_{m,n}} - \sum\limits_{k=0}^{b_{m,n}}\frac{(mb_{m,n})!^{n+1}}{(mk)!^{n+1}}$

$x_{m,n} \in \mathbb{Z}$ as the first term of the previous expression is an integer, and every term of the series is an integer too, as $k \leq b_{m,n}$ for each term.

Let's prove that $x_{m,n} > 0$ by inserting the corresponding series in place of $\frac{a_{m,n}}{b_{m,n}}$:

$x_{m,n} = (mb_{m,n})!^{n+1}\left(\sum\limits_{k=0}^{\infty}\frac{1}{(mk)!^{n+1}} - \sum\limits_{k=0}^{b_{m,n}}\frac{1}{(mk)!^{n+1}}\right) = \sum\limits_{k=b_{m,n}+1}^{\infty}\frac{(mb_{m,n})!^{n+1}}{(mk)!^{n+1}}$

Each term of the series is strictly positive.

Let's prove that $x_{m,n} < 1$, by first showing that, for all terms $k \geq b_{m,n}+1$, we have the upper estimate

$\frac{(mb_{m,n})!^{n+1}}{(mk)!^{n+1}} = \left(\frac{(mb_{m,n})!}{(mk)!}\right)^{n+1} \leq \left(\frac{1}{(mb_{m,n}+1)^{mk-b_{m,n}}}\right)^{n+1}$

Using the formula for infinite geometric series, we have

$x_{m,n} = \sum\limits_{k=b_{m,n}+1}^{\infty}\frac{(mb_{m,n})!^{n+1}}{(mk)!^{n+1}} = \sum\limits_{k=b_{m,n}+1}^{\infty}\left(\frac{(mb_{m,n})!}{(mk)!}\right)^{n+1} < \sum\limits_{k=b_{m,n}+1}^{\infty}\left(\frac{1}{(mb_{m,n}+1)^{mk-b_{m,n}}}\right)^{n+1} = \frac{1}{(mb_{m,n}+1)^{(n+1)(mb_{m,n}-b_{m,n})}((mb_{m,n}+1)^{m(n+1)}-1)}$

$(mb_{m,n}+1)^{(n+1)(mb_{m,n}-b_{m,n})} \geq 1$, because $n+1 \geq 1$ and $mb_{m,n}-b_{m,n} \geq 0$, while $(mb_{m,n}+1)^{m(n+1)}-1 \geq b_{m,n}$ because $m(n+1) \geq 1$. Follows that $x_{m,n} < 1$.

Because $x_{m,n} \in \mathbb{Z}$ and $0 < x_{m,n} < 1$, we have reached a contradiction, so $A(m,n)$ is irrational.

Now, I don't know which criterion Wolfram Alpha uses to test the irrationality of a number, but I'm pretty sure my proof is correct, so I guess this is just an error on their front?