Proof of a Levi-Civita symbol equation How I proof that this equation is true?
$$ \boxed{(-1)^{n-1} \varepsilon_{j_1 ... j_n} x_1^{j_1} \cdots x_{n-1}^{j_{n-1}} x^{j_n}_i = \delta_{in}}\tag{1}$$
where $\varepsilon_{j_1 .. j_n}$ is the completely antisymmetric pseudo-tensor of Levi-Civita, $n \in \mathbb{N}$, and the Einstein summation convention goes from $0$ to $n$ for $i, j_1, ..., j_n$. We define $\textbf{x}_i\equiv (x_i^1, ...,x_i^j , ..., x_i^n)$. I want to demostrate that $\textbf{x}_n$ is orthonormal to an orthonormal set of vectors $\{ \textbf{x}_1, ... , \textbf{x}_{n-1}\}$ in $\mathbb{R}^n$, so this vector has to be
$$\textbf{x}_n =  \begin{vmatrix} \textbf{e}_1 &\cdots &\textbf{e}_n \\ x_1^1 & \cdots & x_1^n \\ \vdots & \ddots & \vdots \\ x_{n-1}^1 & \cdots & x_{n-1}^n\end{vmatrix} \qquad\Leftrightarrow\qquad x_n^i = \varepsilon^{i}_{j_1 \cdots j_{n-1}} x_1^{j_1} \cdots x_{n-1}^{j_{n-1}} . $$
Note that $\{\textbf{e}_1, ..., \textbf{e}_n \}$ is the standard basis of $\mathbb{R}^n$.
For example, for $n=3$
$$ \varepsilon_{abc} x_1^ax_1^bx_2^c = \delta_{13}=0,$$
$$ \varepsilon_{abc} x_2^a x_1^b x_2^c = \delta_{23}=0,$$
$$ \varepsilon_{abc} x_3^ax_1^b x_2^c=\delta_{33}=1,$$
so, the first one is
$$\varepsilon_{abc} x_1^ax_1^bx_2^c = \varepsilon_{1bc}x_1^1x_1^bx_2^c+\varepsilon_{2bc} x_1^2 x_1^b x_2^c + \varepsilon_{3bc}x_1^3 x_1^b x_2^c=$$
$$= \varepsilon_{123}x_1^1x_1^2x_2^3+\varepsilon_{132}x_1^1x_1^3 x_2^2+\varepsilon_{213}x_1^2x_1^1x_2^3 +\varepsilon_{231}x_1^2x_1^3x_2^1 + \varepsilon_{312}x_1^3x_1^1x_2^2+\varepsilon_{321}x_1^3x_1^2x_2^1=$$
$$=x_1^1x_1^2x_2^3-x_1^1x_1^3x_2^2-x_1^2x_1^1x_2^3+x_1^2x_1^3x_2^1+x_1^3x_1^1x_2^2-x_1^3x_1^2x_2^1=0. $$
 A: In case $i \ne n$ then without loss of generality we can assume $i=1$. Then the LHS looks like $(-1)^{n-1} \varepsilon_{j_1 ... j_n} x_1^{j_1} \cdots x_{n-1}^{j_{n-1}} x^{j_n}_1$. We can ignore the sign since we're trying to prove the LHS is zero. Observe if we treat $j$ as a permutation of $\{1,2,\ldots, n\}$ like $i \mapsto j_i$ then the value $\varepsilon_{j_1 ... j_n}$ is just the sign of the permutation $j$. This lets us write the LHS as the sum over permutations. . . .
$\displaystyle \sum_{\sigma \in S_n} $sgn$(\sigma) x_1^{\sigma(1)} \cdots x_{n-1}^{\sigma{(n-1)}} x^{\sigma(n)}_1$
But this is just the determinant of a matrix whose first and last columns are identical. Such a matrix has zero determinant because its columns cannot be a basis for $\mathbb R^n$.
Things become hairier in case $i = n$. Observe the LHS is linear in each input vector and the RHS is not. We can make the LHS arbitrarily large by rescaling the vectors but the RHS will never be greater than $1$. 
So you need some restriction on the vectors $x_1,x_2, \ldots, x_n$ for the equation to hold. For example you could demand $x_1, \ldots x_n$ is an orthogonal basis of unit vectors. In this case the RHS reduces to. . . 
$(-1)^{n-1}\displaystyle \sum_{\sigma \in S_n} $sgn$(\sigma) x_1^{\sigma(1)} \cdots x_{n-1}^{\sigma{(n-1)}} x^{\sigma(n)}_n$
. . . which is very nearly the determinant of an orthogonal matrix and so equal to 1.
