I've stumbled across the family of polynomials
$ f\_p(x) = x^{p-1} + 2 x^{p-2} + \cdots + (p-1) x + p $,
where $p$ is an odd prime. It's not too hard to show that $f\_p(x)$ is irreducible over $\mathbb{Q}$ -- look at the Newton polygon of $f\_p(x+1)$ over $\mathbb{Q}\_p$ and you see that it factors as the product of an irreducible polynomial of degree $p-2$ and a linear. Since $f\_p(x)$ has no real roots (look at the derivative of $f\_p(x) (x-1)^2$) it must be irreducible over $\mathbb{Q}$. It's also not hard to see that the only primes dividing the discriminant are $2, p$ and primes dividing $p+1$. I would expect that the Galois group of a randomish polynomial would be the full symmetric group. Indeed, according to Magma this is true for $f\_p(x)$ for $p=3,5, \dots, 61$ with the exception of $p=7,17$. So my question is -- are these the only exceptions?


Added later: I've had Magma find the Galois group for primes through 101 and it found another exception: $p=97$. So my initial guess was wrong.


Another addition: If one looks at odd $p$ (not just prime) for $p < 100$ there is another exception, 49. Also 241 is  not  an exception (misread magma's output).


The ideas in the following two papers may be of help:


"On the Galois Groups of the exponential Taylor polynomials" by Robert Coleman, in L'Enseignement Mathematique, v 33 (1987) pp 183-189


and


"On the Galois Group of generalized Laguerre polynomials" by Farshid Hajir, J. Th´eor. Nombres Bordeaux 17 (2005), no. 2, 517–525 (also available on the author's web page).



Galois Groups of a family of polynomials
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.


