Whether it be for multivariate cryptography, universal hash functions, substitution boxes, or LFSRs, the irreducible polynomials used for $n$ bits of data always have the $x^n$ and $1$ (a.k.a. $x^0$) terms. The only variance is in which of terms $x^{n-1}$ through $x^1$ are present. And, a multi-bit element $\mathbf{a}=a_{n-1} \parallel \ldots \parallel a_0$ in a finite field is representative of the polynomial $\sum_{i=0}^{n-1} [a_i \cdot x^i]$—that's an $(n-1)$th degree polynomial.
My lack of understanding is what the $x^n$ and $1$ terms correspond to, and the mismatch between the number of bits and degree of the irreducible polynomial used. In an LFSR, for instance, is it $x^n$ or $1$ that does not correspond to a register in the LFSR? Or in other applications of finite fields, such as calculating Rijndael's S-box, what does $x^n$ correspomd to? Is the idea that, where $p_1$ through $p_{n-1}$ are the coefficients, the irreducible polynomial in equation form is $x^n = \sum_{i=1}^{n-1} [p_i \cdot x^i] + 1$? That the $n$th-degree term in the polynomial is separate from the rest of the polynomial?
