Let $\alpha$ be a root of a polynomial $f(x) \in \mathbf{Q}[x]$ of degree $n$, let $K = \mathbf{Q}(\alpha)$, $L$ be the Galois closure of $K$, and $G = \mathrm{Gal}(L/\mathbf{Q}) \subset S_n$. How does one prove that a permutation group contains $A_n$? Following Jordan, the usual method is to show that it is sufficiently highly transitive. Also following Jordan, to do this it suffices to construct subgroups of $G$ which act faithfully and transitively on $n-k$ points and trivially on the other $k$ points (for $k$ large, $\ge 6$ using CFSG), and to show that $G$ is primitive. (The standard method for doing this is to find $l$-cycles for a prime $l$.) In the context of a Galois group, the most obvious place to look for "elements" is to consider the decomposition groups $D$ at places of $\mathbf{Q}$. If $l$ is unramified in $K$, this corresponds to looking at a Frobenius element (conjugacy class). In practice (as far as a computation goes) this is quite useful, but theoretically it is not so great unless there is a prime $l$ for which the factorization is particularly clean. This leaves the places which ramify in $L$. For example, if $v = \infty$, one is considering the action of complex conjugation; if there are exactly two complex roots then $c$ is a $2$-cycle, and from Jordan's theorem (easy in this case) we see that if $G$ is primitive then $G$ is $S_n$.

The proposed method (following Coleman et. al.) for proving that $G$ contains $A_n$ is somewhat misguided, I think. The key point about the polynomial $\sum_{k=0}^{n} x^k/k!$ is that the corresponding field is ramified at many primes, and the decomposition groups at these primes give the requisite elements. Conversely, the polynomial considered in this problem corresponds to a field with somewhat limited ramification - as has been noted, the only primes which ramify divide $p(p+1)$.

It can be hard to compute Galois groups of random families of polynomials in general. I do not know if this is true in the present case, but given the lack of motivation I won't spend any more time thinking about it than the last hour or two, and instead give some partial results. However, the methods given here may well apply more generally. Let $n = p - 1$.

CLAIM: Suppose that $p+1$ is exactly divisible by a prime $l > 3$. Then $G$ contains $A_{n}$. (This applies to a set $p$ of relative density one inside the primes.)

STEP I: Factorization of $p$; $G$ is primitive. Let $f(x) = x^{p-1} + 2 x^{p-2} + \ldots + p$. Note that $$(x-1)^2 f(x) = x(x^{p} - 1) - p(x-1) = x^{p+1} - 1 - (p+1)(x-1).$$ We deduce that $f(x) \equiv x(x-1)^{p-2} \mod p$, and that $$p = \mathfrak{p} \mathfrak{q}^{p-2}$$ for primes $\mathfrak{p}$ and $\mathfrak{q}$ in the ring of integers $O_K$ of $K$ both of norm $p$. (To show this one needs to check that $[O_K:\mathbf{Z}[\alpha]]$ is co-prime to $p$ - one can do this by considering the Newton Polygon of $f(x+1)$.) Let $D \subset G$ be a decomposition group at $p$. This corresponds to choosing a simultaneous embedding of the roots of $f(x)$ into an algebraic closure of the $p$-adic numbers. We see that we may write $f(x) = a(x) b(x)$ as polynomials over the $p$-adic numbers (which I can't latex at this point for some reason), where $a(x) \equiv x \mod p$ has degree one and $b(x) \equiv (x-1)^{p-2}$ is irreducible of degree $p-2$ and corresponds to a totally ramified extension. Clearly $D$ acts transitively on the $p-2 = n-1$ roots of $b(x)$ and fixes the roots of $a(x)$. Since $D \subset G \cap S_{n-1}$, we see that $G \cap S_{n-1}$ is transitive in $S_{n-1}$ and so $G$ is $2$-transitive (and hence primitive).

Step II: Factorization of $l$: Let $l$ be a prime dividing $p+1$. We assume that $l \ge 5$ and $l$ exactly divides $p+1$. We see that $$f(x) \equiv (x-1)^{l-2} \prod_{i=1}^{k-1} (x-\zeta^i)^{l}$$ where $\zeta$ is a $k$th root of unity and $kl=p+1$. This suggests that: $$l = \mathfrak{p}^{l-2} \prod_{i=1}^{k} \mathfrak{q}^l.$$ This also follows from a Newton polygon argument applied to $f(x - \zeta^i)$. (Warning, this uses that $l$ exactly divides $p+1$.)

Step III: Some basic facts about local extensions:

Lemma 1. Suppose the ramification degree of $E/\mathbb{Q}_l$ is $l^m$. Then the ramification degree of the Galois closure of $E$ is only divisible by primes dividing $l(l^m-1)$. Proof. Kummer Theory.

Lemma 2. Suppose that $h(x) \in \mathbf{Q}_l[x]$ is an irreducible polynomial of degree $k$ with $(k,l) = 1$, such that the corresponding field $E/\mathbf{Q}_l$ is totally ramified. If $F$ is the splitting field of $h(x)$, then $\mathrm{Gal}(F/\mathbf{Q}_l) \subset S_k$ contains a $k$-cycle. Proof: From a classification of tamely ramified extensions, there exists an unramified extension $A$ such that $[EA:A] = [E:\mathbf{Q}_l]$ and $EA/A$ is cyclic and Galois. It follows that $\mathrm{Gal}(EA/A)$ acts transitively and faithfully on the roots of $h(x)$, and is thus generated by a $k$-cycle.

Step IV: $G$ contains an $l-2$-cycle. Consider the decomposition group $D$ at $l$. The orbits of $D$ correspond to the factorization of $l$ in $O_K$. On the factors corresponding to primes of the form $\mathfrak{q}^p_i$, the image of $D$ factors through a group whose inertia has degree divisible only by primes dividing $l(l-1)$, by Lemma 1. On the other hand, on the factor corresponding to $\mathfrak{p}^{l-2}$, the image of inertia contains an $l-2$ cycle, by Lemma 2. Since $(l(l-1),l-2) = 1$, we see that $D \subset G$ contains an $l-2$ cycle.

Step V: Jordan's Theorem. Since $G$ is primitive, and $G$ contains a subgroup that acts transitively and faithfully on $l-2$ points (and trivially on all other points), we deduce (from the standard proof of Jordan's theorem) that $G$ is $n-(l-2)+1 = n+3-l$ transitive. This is at least $6$ (since $n+2$ is at least $2l$) and so $G$ contains $A_n$ (by CFSG).

STEP VI: (for you, dear reader) Find the analogous argument when $p+1$ is exactly divisible by $l^k$ for some $k \ge 2$ --- try to construct a cycle of degree $l^k - 2$, although be careful as it will no longer be the case (as it was above) that $[O_K:\mathbf{Z}[\alpha]]$ was co-prime to $l$. This still leaves $p-1$ either a power of $2$ or a power of $2$ times $3$, which might be annoying --- one would have to think hard about the structure of the decomposition group at $2$ in those cases.