In the AKS algorithm, we consider the congruence
$$(X+a)^n \equiv X^n + a \pmod{X^r-1}$$
in $(\mathbb{Z}/n\mathbb{Z})[X]$. The motivation for this is that when taking $r$ in the order of $\log n$, we will always be dealing with polynomials in the order of $\log n$.
It is argued that we take $X^r-1$ because of its simplicity. Why wouldn't we consider the congruence modulo $X^r$, which perhaps could be seen as an even simpler polynomial. Is there an intuitive answer for this?
In the original paper, in a series of statements to prove the algorithm can't label a composite number a 'prime', they use a few properties of cyclotomic polynomials after Lemma 4.6. In particular, they define a group $\mathcal{G}$ which would definitely not be a group if we take $h(X)$ a factor of $X^r$ instead of $X^r - 1$: as $h$ has repeated factors (unless $\deg h = 1$, but that case is trivial), the quotient ring $F_p[X]/(h(X))$ they allude to would contain nontrivial nilpotents.