Let $Y$ be an algebraic variety over a field $K$ whose coordinate ring is given by $K[Y]=K[X_1,...,X_n]/(F)$, where $F=\prod_{i=1}^{n} X_{i}^{a_i}1$ is an element in the polynomial ring $K[X_1,...,X_n]$, with each $a_i$ an integer. Now if we assume that the greatest common divisor $gcd(a_1,...,a_n)=1$, then I want to know that whether the variety $Y$ is irreducible.
There is nothing wrong with Qing's proof of irreducibility but the following is perhaps more conceptual (though less elementary). We immediately reduce to the case when all the $a_i$ are different from zero. Then in the quotient ring, the images of the variables are invertible so we may replace the polynomial ring with the ring of Laurent polynomials whose spectrum is an algebraic torus. Then $F$ is the equation for an algebraic subgroup which is the kernel $K$ of the homomorphism $\mathbb G_m^n \to \mathbb G_m$ given by $(r_1,\ldots,r_n)\mapsto r_1^{a_1}\cdots r_n^{a_n}$. This kernel is a group of multiplicative type whose character group is the cokernel of $\mathbb Z\to \mathbb Z^n$ given by $1\mapsto (a_1,\ldots,a_n)$. The character group of $K$ is the cokernel of this map which is torsion free as the $a_i$ are relatively prime. However, a group of multiplicative type with torsion free character group is a torus and hence irreducible. 


You want to show that $F$ is an irreducible polynomial. If it decomposes as a product $GH$, it is easy to see that at least one variable appears in both $G$ and $H$. Suppose this is $x_1$. Then $F$ is reducible as polynomial in $K(x_2,\dots, x_n)[x_1]$, and the same is true for $x_1^{a_1}f$ where $f=(x_2^{a_2}\cdots x_n^{a_n})^{1}$. Then apply the usual irreducibility criterion for polynomials of the form $X^nb$ (see Lang, Algebra, 3rd edition, VI, §9) and you get a contradiction. 

