Showing that $f(x) = \sum_{k=0}^\infty{} \frac{\cos(k!x)}{(k!)^k}$ is not real analytic for any $x\in\mathbb{R}$

Question

I need to show that$$f(x) = \sum_{k=0}^\infty{} \frac{\cos(k!x)}{(k!)^k}$$ isn`t real analytic. I have shown so far that $f(x)$ is $n$ times differentiable but I struggle with proofing the following: Is $a\in2\pi\mathbb{Q}$ then there is for every $c\gt0$ a $n=n(c)\in\mathbb{N}$ that $f^{(4n)}(a)\gt(4n!)c^{4n}$. My idea was that I can find a big enough $k$ where $k!a$ is always an even multiple of $\pi$ so the $\cos{(k!a)}$ is always equal to 1. So if $a=2\pi\frac{p}{q}$: $$f^{(4n)}(a)=\sum_{k=0}^\infty(k!)^{4n-k}\cos{(k!a)}\gt\sum_{k=q+2}^\infty(k!)^{4n-k}$$ But now I'm stuck. Does anyone have a small hint for me?