Let matrix-valued function $\mathrm M_1 : \mathbb R \to \mathbb R^{2 \times 2}$ be defined as follows
$$\mathrm M_1 (x) := \begin{bmatrix} 1 & x\\ x & 1\end{bmatrix}$$
and let matrix-valued function $\mathrm M_n : \mathbb R \to \mathbb R^{(n+1) \times (n+1)}$ be defined by
$$\mathrm M_n (x) := \begin{bmatrix} \mathrm M_{n-1} (x) & \mathrm v_{n} (x)\\ \mathrm v_{n}^\top (x) & 1\end{bmatrix}$$
where $\mathrm v_{n}^\top (x) := \begin{bmatrix} x^n & \cdots & x^2 & x\end{bmatrix}$. Let function $f_n : \mathbb R \to \mathbb R$ be defined by
$$f_n (x) := \det \mathrm M_n (x) = \det \begin{bmatrix} \mathrm M_{n-1} (x) & \mathrm v_{n} (x)\\ \mathrm v_{n}^\top (x) & 1\end{bmatrix} = \det \left( \mathrm M_{n-1} (x) - \mathrm v_{n} (x) \, \mathrm v_{n}^\top (x) \right)$$
Using the matrix determinant lemma,
$$f_n (x) = \underbrace{\det \left( \mathrm M_{n-1} (x) \right)}_{= f_{n-1} (x)} \cdot \left( 1 - \mathrm v_{n}^\top (x) \, \mathrm M_{n-1}^{-1} (x) \, \mathrm v_{n} (x) \right)$$
Let $\mathrm y (x) := \mathrm M_{n-1}^{-1} (x) \, \mathrm v_{n} (x)$ be the solution of the linear system $\mathrm M_{n-1} (x) \,\mathrm y (x) = \mathrm v_{n} (x)$. Since $\mathrm v_{n} (x)$ is equal to the $n$-th column of $\mathrm M_{n-1} (x)$ multiplied by $x$, the solution is $\mathrm y (x) = x \, \mathrm e_n$. Thus,
$$f_n (x) = f_{n-1} (x) \cdot \left( 1 - \mathrm v_{n}^\top (x) \, \mathrm y (x) \right) = f_{n-1} (x) \cdot \left( 1 - x^2 \right)$$
Since $f_1 (x) = 1 - x^2$, we obtain $\color{blue}{f_n (x) = (1-x^2)^n}$, as found by José Carlos Santos via other means.