When is $X_1^{a_1} \cdots X_n^{a_n}-1$ irreducible? 
Let $F$ be a field, and $a_1, ... , a_n \geq 1$ integers. When is the polynomial $$f = X_1^{a_1} \cdots X_n^{a_n}-1$$  irreducible in $F[X_1, ... ,X_n]$?  

I believe this should be the case if and only if $d = \gcd(a_1, ... , a_n) = 1$.  At least for $F = \mathbb{C}$, the examples I've computed indicate this to be true.
If $d > 1$, then $f$ is not irreducible, since $$f = (X_1^{a_1/d} \cdots X_n^{a_n/d} - 1)[\sum\limits_{i=0}^{d-1} (X_1^{a_1/d} \cdots X_n^{a_n/d})^i]$$
I haven't been able to show the converse yet.
 A: Thanks to darij grinberg for the following approach.  Here are the details written out:
(1): Let $R = F[X_1, ... , X_n], S = \{ (X_1 \cdots X_n)^t : t \geq 0 \}$.  Then $f$ is irreducible in $R$ if and only if it is irreducible in $S^{-1}R = F[X_1, X_1^{-1}, ... , X_n, X_n^{-1}]$.
Since $R$ is a Noetherian unique factorization domain, so is $S^{-1}R$.  If $f$ is irreducible in $R$, then it generates a prime ideal $Rf$, and the extension of this ideal to $S^{-1}R$ remains prime, because $Rf \cap S$ is empty.
Conversely, if $f$ is irreducible in $S^{-1}R$, then it generates a prime ideal $\mathfrak p$ of $S^{-1}R$, because $S^{-1}R$ is a unique factorization domain.  For the inclusion map $R \rightarrow S^{-1}R$, $Rf$ is the contraction of its extension to $S^{-1}R$.  This extension is $\mathfrak p$, so $Rf$ is a prime ideal of $R$, hence $f$ is irreducible.
(2): If $A = (\alpha_{ij}) \in \textrm{GL}_n(\mathbb{Z})$, then $A$ induces a bijection $\mathbb{Z}^n \rightarrow \mathbb{Z}^n$, which induces an $F$-linear automorphism $\phi$ of $S^{-1}R$, because $X_1^{a_1} \cdots X_n^{a_n} : a_i \in \mathbb{Z}$ is an $F$-basis for $S^{-1}R$.  The only thing that isn't clear is that this is actually a ring homomorphism.  To show this, it suffices to show that $\phi$ preserves multiplication of monomials.  But this follows from the fact that $A(v+w) = Av+Aw$, where $v, w \in \mathbb{Z}^n$.
(3): If $a_1, ... , a_n$ are relatively prime integers, there is an $A \in \textrm{GL}_n(\mathbb{Z})$ which maps $(a_1, ... , a_n)$ to $(1, 0, ... , 0)$.  
See this answer Information about Problem. Let $a_1,\cdots,a_n\in\mathbb{Z}$ with $\gcd(a_1,\cdots,a_n)=1$. Then there exists a $n\times n$ matrix $A$ ...
(4): The converse I mentioned in my question: since the GCD $d$ is equal to $1$, we can find an $A$ as in (3), hence a ring isomorphism $\phi$ which maps $f$ to $X_1 - 1$.  Since $X_1 - 1$ is irreducible in $R$, it is so in $S^{-1}R$, hence $f$ is irreducible in $S^{-1}R$, hence in $R$.
A: This is true, and can be seen easily using the Newton polytope of the polynomial and Minkowski sum of polytopes, look for this.
Try to prove that if $F=G*H$, then all the exponents $(b_1,...,b_n)$ of $G$ and $H$ must belong to the segment joining $(0,...,0)$ and $(a_1,...,a_n)$. 
Hint: Prove that


*

*Both $G$ and $H$ have nonzero constant term.

*If $G$ has a monomial that does not belong to the segment $\overline{(0,\ldots,0)(a_1,\ldots,a_n)}$ then $F$ has at least a monomial that is not in the segment $\overline{(0,\ldots,0)(a_1,\ldots,a_n)}$.

