using vandermonde determinant to express general determinant? I am reading an introduction paragraph about Vandermonde determinant and the following statement troubles me:
[Statement]: For a determinant
$$D_n=\begin{vmatrix} x_{1n}&x_{1,n-1}&...&x_{11}\\x_{2n}&x_{2,n-1}&...&x_{21}\\...&...&...&...\\x_{nn}&x_{n,n-1}&...&x_{n1}\end{vmatrix}$$
Suppose that in a purely formal (?) way we replace each element $x_{ik}$ by $x_{i}^{k-1}$. After we have made such a substitution, the determinant above clearly becomes the Vandermonde determinant. This leads at once to the formula
$$D_n=\begin{align}(x_1-x_2)(x_1-x_3)\ ...\ (x_1-x_n)\\\times(x_2-x_3)\ ...\ (x_2-x_n)\\............\\\times(x_1-x_n)\end{align}$$
Multiply out the terms in $D_n$ above, and in each of the products so obtained, replace $x_k^{s-1}$ by $x_{ks}$. If a product does not contain some power of the number $x_k$, then supply the factor $x_k^0$, which after substitution becomes $x_{k1}$. We note that this rule can be adopted as the definition of a determinant (?)
[Questions]:

*

*I don't get how can this be the definition of a determinant, can someone kindly visualize it for me ?

*Is there any limitation for this kind of substitution? If not, what is the justification?

 A: 1.   It's easier to follow on a simple case, for example a $3 \times 3$ determinant with the known expansion:
$$
\begin{vmatrix}
   a_{3} &a_{2} &a_{1}
\\ b_{3} &b_{2} &b_{1}
\\ c_{3} &c_{2} &c_{1}
\end{vmatrix}
 \;\;=\;\; -a_1 b_2 c_3 + a_1 b_3 c_2 + a_2 b_1 c_3 - a_2 b_3 c_1 - a_3 b_1 c_2 + a_3 b_2 c_1 \tag{1}
$$
Making the substitutions $a_3 \to a^2, a_2 \to a^1, a_1 \to a^0 \dots\,$ and using the Vandermonde expression:
$$
\begin{vmatrix}
   a^{2} &a &1
\\ b^{2} &b &1
\\ c^{2} &c &1
\end{vmatrix}
 \;\;=\;\; (a-b)(a-c)(b-c) \tag{2}
$$
Expanding the product in $(2)$:
$$
\begin{align}
(a-b)(a-c)(b-c) \;&=\; a^2 b - a^2 c - a b^2 + a c^2 + b^2 c - b c^2
\\ \;&=\; a^2 b^1c^0 - a^2 b^0 c^1 - a^1 b^2 c^0 + a^1 b^0 c^2 + a^0 b^2 c^1 - a^0 b^1 c^2 \tag{3}
\end{align}
$$
Finally, making the reverse substitutions $a^2 \to a_3, a^1 \to a_2, a^0 \to a_1$ in $(3)$ turns out to give the same expression as the right hand side of $(1)$.
The "rule" can be formally proved to work in the general case, though calling it a possible definition of the determinant is a bit of a stretch.
2.   The substitution can be justified in that very particular context. Without pretense of a formal proof, but the idea is that the substitutions $x_{ik} \to x_i^{k-1}$ are reversible in that case, because each term of the determinant is a monomial containing exactly one element from each line and each column, so each term in $(3)$ can be univocally traced back to the elements in the matrix whose product was taken. For example, the first term $a^2b$ can only be the product of the $a^2$ in the top left corner, the $b$ in the middle, and the $1$ at the intersection of the remaining row and column, which is the bottom right corner. Because of the reversibility of the substitution, it is possible to translate the problem "purely formally" into the other one with a known result, then transfer that result back to the original problem.
Using such "trick" substitutions must be done very carefully, lest they produce bogus results, or just plain fail. Suppose for example that you tried to mechanically apply the same argument to calculate the following:
$$
\begin{vmatrix}
   a_2 &a_1
\\ a_1 &a_2
\end{vmatrix}
\;\;\to\;\; 
\begin{vmatrix}
   a^1 &a^0
\\ a^0 &a^1
\end{vmatrix}
\;\;=\;\;
\begin{vmatrix}
   a &1
\\ 1 &a
\end{vmatrix}
\;\;\to\;\;
a^2 - 1
\;\;\to\;\;
\text{???}
$$

[ EDIT ]   What the quoted statement does is essentially the following, where the blue arrows denote the substitutions from and back to the $\left(x_{ij}\right)$ space, and the red arrows indicate algebraic manipulations in the other $\left(x_i^{j-1}\right)$ space.
$$\displaystyle \left|x_{ij}\right| \color{blue}{\to} \left|x_i^{j-1}\right| \color{red}{\to} \prod_{i \lt j} (x_i-x_j) \color{red}{\to} \sum _{\sigma \in S_{n}}\left(\operatorname {sgn}(\sigma )\prod _{i=1}^{n}x_{i}^{\sigma _{i}-1}\right) \color{blue}{\to} \sum _{\sigma \in S_{n}}\left(\operatorname {sgn}(\sigma )\prod _{i=1}^{n}x_{i,\sigma _{i}}\right)$$
This says that, assuming (a) the determinant is multilinear of a certain form, and (b) the Vandermonde formula holds true in the particular case of a Vandermonde determinant, then the general determinant formula can be univocally derived. It goes without saying that this is a theoretical result, not a practical "trick", and not a method to calculate specific determinants.
A: To calculate the determinant of the matrix
$(x_{ij})_{i,j=1,n}$,  switch over to the determinant of $(x_i^j)_{i,j=1,n}$, identify the latter as the product
$$x_1 \cdots x_n \cdot \prod_{i<j} (x_j - x_i)$$
expand into a sum
$$\sum \pm x_1^{\alpha_1} \cdots x_n ^{\alpha_n}$$
now switch back from exponents to second indexes and get the value of the original determinant
$$\sum \pm x_{1 \alpha_1} \cdots x_{n \alpha_n}$$
This was the original idea of Vandermonde for determinants, later explained by Cauchy, see this.
