Irreducibility of $f(x)=x^n-p^m$ I have a question.
Let $f(x)=x^n-p^m$ where $m$ and $n$ are coprime and $p$ is prime integer.
How can I show that $f(x)$ is irreducible over $\Bbb Q$?
 A: Hats off to Lord Shark the Unknown for providing the key that unlocks the puzzle-door on this one; taking a tip from his comment on the question, I have tried below to flesh out an answer.  I wanted to see how things work out in detail, so . . . 
First of all, let's consider the case 
$\gcd(n, m) = d > 1; \tag 1$
then since
$d \mid m, \; d \mid n, \tag 2$
we may write
$n = n_1 d, \; m = m_1 d, \tag 3$
and we have
$X^n - p^m = X^{n_1 d} - p^{m_1 d} = (X^{n_1})^d - (p^{m_1})^d$
$= (X^{n_1} - p^{m_1}) \displaystyle \sum_0^{d - 1} (X^{n_1})^{d - 1 - i} (p^{m_1})^i, \tag4$
which demonstrates that $X^n - p^m$ is in fact reducible in the event that $\gcd(n, m) > 1$; thus the condition $\gcd(n, m) = 1$ is not in fact inconsequential, insofar as does distinguish 'twixt reducible and irreducible cases of $X^n - p^m$, provided of course the question of the problem is answerable in the affirmative, which we shall indeed herein establish.
We begin with $m = 1$; here we trivially have $\gcd(n, m) = 1$, and the Eisenstein criterion with prime $p$ directly applies to show $X^n - p$ is irreducible over $\Bbb Q$.  However, Eisenstein does not apply to $X^n - p^m$ when $m \ge 2$; therefore we seek another line of analysis. 
We consider the field extension of $\Bbb Q$ formed by adjoining $\sqrt [n] p$; that is, the field $\Bbb Q(\sqrt [n] p)$; since 
$X^n - p$ is irreducible over $\Bbb Q$, we have from elementary considerations that 
$\Bbb Q(\sqrt[n] p) \simeq \Bbb Q[X] / \langle X^n - p \rangle, \tag 5$
where
$\langle X^n - p \rangle = (X^n - p) \Bbb Q[X]  \tag 6$
is the principal ideal generated by $X^n - p$ in $\Bbb Q[x]$.  Furthermore it follows from (5) that
$[\Bbb Q(\sqrt[n] p): \Bbb Q] = n; \tag 7$
that is, the dimension of $\Bbb Q(\sqrt[n] p)$ as a vector space over $\Bbb Q$ is $n$.  
We next observe that
$\sqrt[n]{p^m} = (\sqrt[n] p)^m \in \Bbb Q(\sqrt[n] p), \tag 8$
from which it is seen that
$\Bbb Q(\sqrt[n] {p^m}) \subset \Bbb Q(\sqrt[n] p); \tag 9$
we now show that under the hypothesis
$\gcd(n, m) = 1 \tag{10}$
we also have
$\sqrt[n] p \in \Bbb Q(\sqrt[n]{p^m}); \tag{11}$
for, as is well-known, (10) implies that there are $a, b \in \Bbb Z$ such that
$an + bm = 1; \tag{12}$
then
$\sqrt[n] p = (\sqrt[n] p)^1 = (\sqrt[n] p)^{an + bm} = (\sqrt[n] p)^{an} (\sqrt[n] p)^{bm}$
$= ((\sqrt[n] p)^n)^a ((\sqrt[n] p)^m)^b = p^a (\sqrt[n] {p^m})^b \in \Bbb Q(\sqrt[n]{p^m}); \tag{13}$
since, then
$\sqrt[n] p \in \Bbb Q(\sqrt[n]{p^m}), \tag{14}$
it follows that
$\Bbb Q(\sqrt[n] p) \subset \Bbb Q(\sqrt[n]{p^m}), \tag{15}$
and thus, via (9),
$\Bbb Q(\sqrt[n] p) = \Bbb Q(\sqrt[n]{p^m}), \tag{16}$
and now from (7) we see that
$[\Bbb Q(\sqrt[n]{p^m}): \Bbb Q] = n \tag{17}$
as well.
Now if $X^n - p^m$, $m > 1$, were reducible over $\Bbb Q$, we could write
$X^n - p^m = r(X) s(X), \tag{18}$
with
$r(X), s(X) \in \Bbb Q[X], 0 < \deg r, \deg s < n, \tag{19}$
then
$r(\sqrt[n]{p^m}) s(\sqrt[n]{p^m}) = (\sqrt[n]{p^m})^n - p^m = p^m - p^m = 0, \tag{20}$
so that at least one of
$r(\sqrt[n]{p^m}), s(\sqrt[n]{p^m}) = 0; \tag{21}$
then in the light of (19), we may conclude that
$[\Bbb Q(\sqrt[n]{p^m}): \Bbb Q] < n, \tag{22}$
since $\sqrt[n]{p^m}$ satisfies a polynomial of degree less than $n$.  This of course stands in contrediction to (17);
therefore, 
$X^n - p^m \tag{23}$
is irreducible over $\Bbb Q$.
A: This is a nice example for using Newton polygons, tool from which criteria such as Eisenstein criterion follow. In this case the Newton polygon of
$$
f(x)=x^n-p^m
$$
with respect to prime $p$ is a line segment with end points $(0,m)$ and $(n,0)$.
Now if there are no other integer points on this line, then by Dumas criterion the $f(x)$ is irreducible in $\mathbb{Q}[x]$ . Since you have $(m,n)=1$, there are no such integer points, and the polynomial is indeed irreducible.
Basically you can translate this into generalized version of Eisenstein's criterion in terms of prime powers, such as:

If $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0 \in \mathbb{Q}[x]$, such that $a_0a_n\neq 0$. Then if there is a prime $p$ such that $p \nmid a_n$, $p^k\mid a_i$ for $i<n$, $p^{k+1} \nmid a_0$, and $(k,n)=1$, then $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Choosing $k=1$ gives you the familiar Eisenstein's criterion, while in your case you need $k=m$.
