I want to demostrate that the vector defined by $$ \textbf{x}_n = \begin{vmatrix} \textbf{e}_1 & \cdots & \textbf{e}_n \\ x_1^1 & \cdots & x_1^n \\ \vdots & \ddots & \vdots \\ x_{n-1}^1 & \cdots & x_{n-1}^n \end{vmatrix}$$ form an orthonormal basis $\{\textbf{x}_i\}_{i=1}^n$ in $\mathbb{R}^n$. I don't want to use Gram-Schmidt method to obtain $\textbf{x}_n$ but I've obtained the vectors $\{\textbf{x}_i\}_{i=1}^{n-1}$ by Gram-Schmidt method. The standard basis in $\mathbb{R}^n$ is $\{\textbf{e}_i \}_{i=1}^n$. The $j-$ component of the $i-$vector $\textbf{x}_i\in \mathbb{R}^n$ is $x_i^j\in \mathbb{R}$.
My way by induction: seen what happen for $n=2,3$, I suppose it's true for $n$ so $$ \begin{vmatrix} x_n^1 & \cdots & x_n^n \\ x_1^1 & \cdots & x_1^n \\ \vdots & \ddots & \vdots \\ x_{n-1}^1 & \cdots & x_{n-1}^n \end{vmatrix}=1 \qquad \Longrightarrow \qquad x_{n+1}^{n+1} + \sum_{i=1}^n x_i^{n+1} \begin{vmatrix} x_1^1 & \cdots & x_1^{i-1} & x_1^{i+1} & \cdots & x_1^{n+1} \\ \vdots & & \vdots& \vdots& &\vdots \\ x_n^1 & \cdots & x^{i-1}_n & x^{i+1}_n & \cdots & x^{n+1}_n \end{vmatrix} \overset{?}{=}1$$