This question is inspired by my answer to the question "How to compute $\prod_{n=1}^\infty\left(1+\frac{1}{n!}\right)$?".

The sums $f(k) = \sum_{n=1}^{\infty} \frac{1}{(n!)^k}$ (for positive integer $k$) came up, and I noticed that $f(1) = e-1$ was transcendental and $f(2) = I_0(2)-1$ (modified Bessel function) was probably transcendental since $J_0(1)$ (Bessel function) is transcendental.

So, I made the conjecture that $all$ the $f(k)$ are transcendental, and I am here presenting it as a question.

The only progress I have made is to show that all the $f(k)$ are irrational.

This follows the standard proof that $e$ is irrational: if $f(k) = \frac{a}{b} =\sum_{n=1}^{\infty} \frac{1}{(n!)^k}$, multiplying by $(b!)^k$ gives $a (b!)^{k-1}(b-1)! =\sum_{n=1}^{b} \frac{(b!)^k}{(n!)^k} +\sum_{n=b+1}^{\infty} \frac{(b!)^k}{(n!)^k} $ and the left side is an integer and the right side is an integer plus a proper fraction (easily proved).

I have not been able to prove anything more, but it somehow seems to me that it should be possible to prove that $f(k)$ is not the root of a polynomial of degree $\le k$.