<s>[INST] How to compute the determinant of this Toeplitz matrix? 
Given a positive integer $n$, express$$
f_n(x) = \left|\begin{array}{c c c c c} 
1 & x & \cdots & x^{n - 1} & x^n\\
x & 1 & x & \cdots & x^{n - 1} \\
\vdots & x & \ddots & \ddots & \vdots\\
x^{n - 1} & \vdots & \ddots & 1 & x\\
x^n & x^{n - 1} & \cdots & x & 1
\end{array}\right|
$$
  as a polynomial of $x$.

I tried to find a recurrence relation of $\{f_n\}_{n \geqslant 1}$ using Laplace expansion, but there seems to be no patterns in the minors in the expansion. Is there a somewhat simple recurrence relation of $\{f_n\}_{n \geqslant 1}$ or these determinants can be computed with other methods?
 [/INST] This is simply an illustration of José Carlos Santos' answer.

Subtracting $x$ times the second column from the first gives
$$
\begin{align}
f_n(x)
&=\det\begin{bmatrix}
1&x&x^2&x^3&\cdots&x^n\\
x&1&x&x^2&\cdots&x^{n-1}\\
x^2&x&1&x&\cdots&x^{n-2}\\
x^3&x^2&x&1&\cdots&x^{n-3}\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
x^n&x^{n-1}&x^{n-2}&x^{n-3}&\cdots&1
\end{bmatrix}\\
&=\det\begin{bmatrix}
\color{#C00}{1-x^2}&x&x^2&x^3&\cdots&x^n\\
0&\color{#090}{1}&\color{#090}{x}&\color{#090}{x^2}&\color{#090}{\cdots}&\color{#090}{x^{n-1}}\\
0&\color{#090}{x}&\color{#090}{1}&\color{#090}{x}&\color{#090}{\cdots}&\color{#090}{x^{n-2}}\\
0&\color{#090}{x^2}&\color{#090}{x}&\color{#090}{1}&\color{#090}{\cdots}&\color{#090}{x^{n-3}}\\
\vdots&\color{#090}{\vdots}&\color{#090}{\vdots}&\color{#090}{\vdots}&\color{#090}{\ddots}&\color{#090}{\vdots}\\
0&\color{#090}{x^{n-1}}&\color{#090}{x^{n-2}}&\color{#090}{x^{n-3}}&\color{#090}{\cdots}&\color{#090}{1}
\end{bmatrix}\\[6pt]
&=\color{#C00}{\left(1-x^2\right)}\color{#090}{f_{n-1}(x)}
\end{align}
$$
Since $f_0(x)=1$, we have that
$$
f_n(x)=\left(1-x^2\right)^n
$$
</s>