So I'm given $f(x) = \sum_{k=0}^{8}\frac{x^k}{k!} \in \mathbb{Q}[x]$. Denote its splitting field by $E$, then I'm also given that ${\rm Gal}(E/\mathbb{Q}) \cong A_8$. The task is to prove that $f(x)$ is irreducible over $\mathbb{Q}$.
So $A_8$ is known to be simple, so by the fundamental theorem there cannot exist intermediate fields $\mathbb{Q} \subset K \subset E$ such that $\mathbb{Q} \subset K$ is normal. So if $f$ is reducible over $\mathbb{Q}$, but not completely, then the splitting field of the remaining irreducible polynomial will also be $E$. Any irreducible polynomial in $\mathbb{Q}[x]$ which splits in $E$ will not split in any subfield of $E$ containing $\mathbb{Q}$.
Assuming that my reasoning so far is correct, I cannot see how to proceed from here. I don't quite see for instance how it can't be the case that one root of $f$ lies in $\mathbb{Q}$ and the rest in $E$. Any help is appreciated.
You can appeal to the fact that $f$ is irreducible iff the Galois group acts transitively on the set of roots. Here, we expect that $\text{Gal}(E/\mathbb{Q}) \cong A_8$ under the action of the Galois group on the roots, but let's ensure that is true:
Consider the action of the Galois group on the $n$ roots of $f$ in $\mathbb{C}$. This induces a homomorphism $f:A_8 \to S_n$. Since $A_8$ is simple, this must be injective, and so $n \geq 8$. Since $n\leq 8$ anyway, $n=8$. So $[S_8:f(A_8)] = 2$, so $f(A_8) \vartriangleleft S_8$, whence $f(A_8) = A_8$. In particular, the Galois group acts transitively on the set of roots of $f$.
So, $f$ is irreducible.