Here is a broader setup for your question. Let $A$ be a Dedekind domain with fraction field $F$, $E/F$ be a finite Galois extension, and $B$ be the integral closure of $A$ in $E$.
Pick a prime $\mathfrak p$ in $A$ and a prime $\mathfrak P$ in $B$ lying over $\mathfrak p$.
The decomposition group $D(\mathfrak P|\mathfrak p)$ naturally maps by reduction mod $\mathfrak P$ to the automorphism group $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$ and Frobenius showed this is surjective. The kernel is the inertia group, so if $\mathfrak p$ is unramified in $B$ then we get an isomorphism from $D(\mathfrak P|\mathfrak p)$ to $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$, whose inverse is an embedding
of the automorphism group of the residue field extension into $\text{Gal}(E/F)$.
If we take $A = {\mathbf Z}$ then we're in the number field situation and this is where Frobenius elements in Galois groups come from.
In your case you want to take $A = {\mathbf Q}[t]$, so $F = {\mathbf Q}(t)$.
You did not give any assumptions about $f(x,t)$ as a polynomial in ${\mathbf Q}[x,t]$. (Stylistic quibble: I think it is better to write the polynomial as $f(t,x)$, specializing the first variable, but I'll use your notation.) Let's assume $f(x,t)$ is absolutely irreducible, so the ring $A' = {\mathbf Q}[x,t]/(f)$ is integrally closed. [EDIT: I should have included the assumption that $f$ is smooth, as otherwise $A'$ will not be integrally closed, but this "global" int. closed business is actually not so important. See comments below.] Write $F'$ for the fraction field of $A'$. After a linear change of variables we can assume $f(x,t)$ has a constant coefficient for the highest power of $x$, so $A'$ is the integral closure of $A$ in $F'$.
Saying for some rational $t_0$ that the specialization $g(x) = f(x,t_0)$ is separable in ${\mathbf Q}[x] = (A/(t-t_0))[x]$ implies the prime $(t-t_0)$ is unramified in $A'$. Let $E$ be the Galois closure of $F'/F$ and $B$ be the integral closure of $A$ in $E$. A prime ideal that is unramified in a finite extension is unramified in the Galois closure, so $(t-t_0)$ is unramified in $B$. For any prime $\mathfrak P$ in $B$ that lies over $(t-t_0)$, the
residue field $B/\mathfrak P$ is a finite extension of $A/(t-t_0) = \mathbf Q$ and since
$E/F$ is Galois the field $B/\mathfrak P$ is normal over $A/(t-t_0)$. These residue fields have characteristic 0, so they're separable: $B/\mathfrak P$ is a finite Galois extension of $\mathbf Q$. I leave it to you to check that $B/\mathfrak P$ is the Galois closure of $g(x) = f(t_0,x)$ over $\mathbf Q$. Then the isomorphism of $D(\mathfrak P|(t-t_0))$ with $\text{Aut}((B/\mathfrak P)/\mathbf Q) = \text{Gal}((B/\mathfrak P)/\mathbf Q)$ provides (by looking at the inverse map) an embedding of the Galois group of $g$ over $\mathbf Q$ into the Galois group of $f(x,t)$ over $F = {\mathbf Q}(t)$.
I agree with Damiano that there are problems when the specialization is not separable. In that case what happens is that the Galois group of the residue field extension is identified not with the decomposition group (a subgroup of the Galois group of $E/F$) but with the quotient group $D/I$ where $I = I(\mathfrak P|\mathfrak p)$ is the inertia group, and you don't generally expect a proper quotient group of a subgroup to naturally embed into the original group.
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.
Best Answer
[Edited mostly to incorporate references etc. from Michael Zieve]
The polynomial $(x^n-1)/(x-1) - y$ has Galois group $S_{n-1}$ over ${\bf C}(y)$ for each $n$, as expected. This answers one of the three problems; the question statements asserts that they are equivalent, which doesn't seem to be the case (see my comment there), so I hope I'm answering the intended one.
The proof combines group theory and polynomial algebra (as might be expect) with algebraic topology, which might be a bit less expected. The connection is as follows. Let $S$ be the Riemann sphere with coordinate $y$, let $P(x,y)$ be any polynomial of degree $d>0$ in $x$, and let $S'$ be the Riemann surface corresponding to $P(x,y)=0$. The rational function $y$ on $S'$ gives a map $S' \rightarrow S$; let $B = \lbrace y_1,\ldots,y_b\rbrace$ be its set of branch points, and fix some $\eta \in S$ not in $B$. Then $\pi_1(S-B,\phantom.\eta)$ is a free group on $b-1$ generators, with generators $g_1,\ldots,g_b$ (where each $g_j$ is a loop from the base point $\eta$ around $y_j$ and back to $\eta$) subject to the single relation $g_1 g_2 \cdots g_b = 1$. These lift to automorphisms $\tilde g_j$ of $S' - y^{-1}(B)$ such that $y = y \cdot \tilde g_j$, and thus to permutations $\pi_j$ of the $d$ sheets of $S'$ above $S$. Then the key fact is that
In our case $P\phantom.$ has the form $p(x) - y$. In general I think it is known which polynomials (or even rational functions) $p$ yield $p(x)-y$ with Galois group other than $S_d$, but it's not easy and involves the classification of finite simple groups! [Actually even less is known; the groups have been determined, but the polynomials, not quite, even in the polynomial case. See below. But this is tangential to the question at hand (albeit a fascinating tangent) because the ensuing argument is enough.] Fortunately in our case the following sufficient condition is all we need:
[Naturally this result is not new, but I was still surprised to learn from Mike Zieve of its true age and pedigree: Hilbert (1892)! See below.]
Proof: Here $b=d$ and the branch points are $y_1,\ldots,y_{d-1}$ and $y_d = \infty$. Since $P$ is a polynomial, $\pi_d$ is a $d$-cycle. (This already proves that the Galois group is transitive, that is, that the polynomial is irreducible; but we already knew this because it is obviously irreducible over ${\bf C}[y]$.) Each $\pi_j$ for $j<d$ is a simple transposition. So we have $d-1$ transpositions whose product is a $d$-cycle. But we readily prove the following by induction on $d$:
Here (i) is satisfied, so (iii) holds and the Galois group is $S_d$ as claimed.
[NB we cannot drop the condition that the $y_j$ be distinct; e.g. the Čebyšev polynomial $T_d$ has dihedral Galois group, which is smaller than $S_d$ once $d>3$, even though $T'_d$ has distinct roots $x_j$, because $T_d(x_j) = \pm 1$ for each of them.]
It remains to check that the hypothesis on $p$ is satisfied when $d=n-1$ and $p=(X^n-1)/(X-1)$. For lack of a better idea I did this by computing the discriminant of the polynomial $$ q(y) = \frac{(n-1)^{n-1} y^n - n^n (y-1)^{n-1}}{(y-n)^2} \qquad (1) $$ of degree $n-2$ whose roots are the $y_j$, and checking that it is nonzero; in fact $$ \mathop{\rm disc} q(y) = \pm 2 \Delta_{n-1} \Delta_n $$ where $\Delta_m := m^{(m-1)(m-3)}$ (and the sign depends on $n \bmod 4$). The formula for $q$ was obtained from the familiar expression $\pm \bigl( (n-1)^{n-1} A^n - n^n B^{n-1} \bigr)$ for the discriminant of the trinomial $x^n - Ax + B$. It follows that the numerator of (1) is $\pm$ the discriminant of $x^n - xy + (y-1) = (x-1) (p(x)-y)$ as a polynomial in $x$, whence (1) soon follows; substituting $y = nz/(nz-(n-1))$, and using the covariance of the discriminant under ${\rm PGL}_2$ together with the same trinomial formula, soon gives the claimed $\pm 2 \Delta_{n-1} \Delta_n$ (the factor of $2$ arises at the end from a double application of L'Hôpital's rule), QED.
Here are some relevant references and additional information.
Polynomials with distinct critical values: This is in section 4.4 of Serre's Topics in Galois Theory (Boston: Jones & Bartlett 1992), as is the reference to Hilbert: "Ueber die irreduzibilität genzen rationalen Funktionen mit ganzzahligen Koeffizienten", J. reine angew. Math. ("Crelle's J.") 110, 104-129 (= Ges. Abh. II, 264-286). Serre gives such polynomials the suggestive name "Morse functions". M. Zieve also notes that a 1959 paper by Birch and Swinnerton-Dyer in Acta Arith. (MR0113844 (22 #4675)) attributes to Davenport the equivalent formulation
They call this a "more euphonious form"; that's a matter of taste, but at any rate it's a memorable formulation, and the one that ended up being used here.
Exceptional Galois groups of $p(x) = y$. The primitive groups that can occur are described by Peter Müller in a paper Primitive monodromy groups of polynomials. They are the symmetric and alternating groups, plus cyclic and dihedral groups (for polynomials equivalent to powers and Čebyšev), and a finite but substantial list of exceptional possibilities. It is hopeless to classify all cases of alternating Galois group. For the rest it can be done, but not easily. Some are exhibited by Cassou-Nogues and Couveignes in an Acta Arith. paper Factorisations explicites de $g(y)-h(z)$; most others were done by Mike Zieve himself; and the final case, polynomials of degree $23$ with Galois group the Mathieu group $M_{23}$, I computed only a few days ago after Mike noted it was still open (they're defined over the quadratic extension of ${\bf Q}(\sqrt{-23})$ of discriminant $3 \cdot 23^3$).
When $p$ is allowed to be a rational function, even the list of groups is not yet known. Mike notes that this "ties in with the 'genus zero program' initiated by Guralnick and Thompson" that was completed in a 2001 paper by Frohardt and Magaard in the Annals of Math. Thanks again to Mike Zieve for all this information.