I am currently trying to understand the concept of a cyclotomic polynomial which, due to my absent knowledge of rings/fields/etc, is proving difficult. From Wikipedia, the following definition is provided:
In mathematics, more specifically in algebra, the $n^\text{th}$ cyclotomic polynomial, for any positive integer $n$, is the unique irreducible polynomial with integer coefficients that is a divisor of $x^{n}-1$ and is not a divisor of $x^{k}-1$ for any $k < n$.
My problem with this definition is that $x^n - 1$ divides $x^n - 1$ but not $x^k - 1$ for $k < n$. Thus, since the $n^\text{th}$ cyclotomic polynomial is unique, this means that it is always $x^n - 1$?
A second definition that I have found, from Wolfram MathWorld, is that the $n^\text{th}$ cyclotomic polynomial is the irreducible polynomial given by $$ \Phi_n (x) = \prod_{k=1}^n \left( x - \zeta_k \right) $$ where $\zeta_k$ denotes the 'primative $k^\text{th}$ root of unity'. However, my understanding of 'irreducible' polynomial is one which cannot be broken into smaller factors. From the definition above, $\Phi_n (x)$ can clearly be factorised, since each $\left( x - \zeta_k \right)$ represents a factor?
For example $x^3-1=(x-1)(x^2+x+1)$, and for $n=3$ the cyclotomic polynomial is $x^2+x+1=0$. That polynomial is irreducible over $\mathbb{Q}$, i.e. it cannot be broken into smaller factors with coefficients in $\mathbb{Q}$, which is what matters for this definition (although, as you note, it is not irreducible over $\mathbb{C}$).
In general, an easy induction argument proves that $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^2+x+1) $$ which answers your question: $x^n-1$ is never irreducible over $\mathbb{Q}$ and so the $n^{\text{th}}$ cyclotomic polynomial is never $x^n-1$.
One must then ponder the general question, for any $n$, whether the polynomial $x^{n-1}+x^{n-2}+...+x^2+x+1$ is irreducible over $\mathbb{Q}$ (it is for $n=3$). But that's another question.