What is the principle to calculate the determinant of Vandermonde matrix with $f_j(t)$? 
Let $$V = \left( \begin{array}{cccc}{1} & {x_{1}} & {\cdots} & {x_{1}^{n-1}} \\ {1} & {x_{2}} & {\cdots} & {x_{2}^{n-1}} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {1} & {x_{n}} & {\cdots} & {x_{n}^{n-1}}\end{array}\right)$$ Compute $\det(V)$.

If we assume: 
$$f_{1}(t)=1, f_{j}(t)=\prod_{i=1}^{j-1}\left(t-x_{i}\right)=\sum_{i=1}^{j} a_{i j} t^{i-1}, \quad 2 \leqslant j \leqslant n$$
Vandermonde Matrix can write into:
$$\left( \begin{array}{cccc}{1} & {x_{1}} & {\cdots} & {x_{1}^{n-1}} \\ {1} & {x_{2}} & {\cdots} & {x_{2}^{n-1}} \\ {\vdots} & {\vdots} & {\cdots} & {\vdots} \\ {1} & {x_{n}} & {\cdots} & {x_{n}^{n-1}}\end{array}\right) \left( \begin{array}{cccc}{1} & {a_{12}} & {\cdots} & {a_{1 n}} \\ {0} & {1} & {\cdots} & {a_{2 n}} \\ {\vdots} & {\ddots} & {\ddots} & {\vdots} \\ {0} & {\cdots} & {0} & {1}\end{array}\right)=\left( \begin{array}{cccc}{f_{1}\left(x_{1}\right)} & {0} & {\cdots} & {0} \\ {f_{1}\left(x_{2}\right)} & {f_{2}\left(x_{2}\right)} & {\ddots} & {\vdots} \\ {\vdots} & {\vdots} & {\ddots} & {0} \\ {f_{1}\left(x_{n}\right)} & {f_{2}\left(x_{n}\right)} & {\cdots} & {f_{n}\left(x_{n}\right)}\end{array}\right)$$
then $det(V)det(I)=det(*)$ because that they all $\in\mathbb{F}^{n\times n}$.
but I cannot understand how to manufacture the element of *, on the other word,  $f_j(t)$.
Why can we set up a function $f_j(t)$? How does this function be assumed?
What is the principle behind this $f_j(t)$?

further more：
$C=\left( \begin{array}{cccc}{\frac{1}{s_{1}-t_{1}}} & {\frac{1}{s_{1}-t_{2}}} & {\cdots} & {\frac{1}{s_{1}-t_{n}}} \\ {\frac{1}{s_{2}-t_{1}}} & {\frac{1}{s_{2}-t_{2}}} & {\dots} & {\frac{1}{s_{2}-t_{n}}} \\ {\vdots} & {\vdots} & {\cdots} & {\vdots} \\ {\frac{1}{s_{n}-t_{1}}} & {\frac{1}{s_{n}-t_{2}}} & {\cdots} & {\frac{1}{s_{n}-t_{n}}}\end{array}\right)=\left(\frac{q_{j}\left(s_{i}\right)}{p\left(s_{i}\right)}\right)=\\\left( \begin{array}{cccc}{\frac{1}{p(s_{1})}} & { } & {  } & {  } \\ {  } & {\frac{1}{p\left(s_{2}\right)}} \\ { } & { } & { \cdots } \\{ }& { }& { }&{\frac{1}{p\left(s_{n}\right)}}\end{array}\right)\left( \begin{array}{cccc}{1} & {s_{1}} & {\cdots} & {s_{1}^{n-1}} \\ {1} & {s_{2}} & {\cdots} & {s_{2}^{n-1}} \\ {\vdots} & {\vdots} & {\cdots} & {\vdots} \\ {1} & {s_{n}} & {\cdots} & {s_{n}^{n-1}}\end{array}\right)\left( \begin{array}{cccc}{1} & {t_{1}} & {\cdots} & {t_{1}^{n-1}} \\ {1} & {t_{2}} & {\cdots} & {t_{2}^{n-1}} \\ {\vdots} & {\vdots} & {\cdots} & {\vdots} \\ {1} & {t_{n}} & {\cdots} & {t_{n}^{n-1}}\end{array}\right)^{-1}\left( \begin{array}{cccc}{q_1(t_1)} & { } & {  } & {  } \\ {  } & {q_2(t_2)} \\ { } & { } & { \cdots } \\{ }& { }& { }&{q_n(t_n)}\end{array}\right)$
Assumed $p(x)=\prod_{i=1}^{n}\left(x-t_{i}\right), q_{j}(x)=\frac{p(x)}{x-t_{j}}=\sum_{i=1}^{n} a_{i j} x^{i-1}​$ 
like $f_j(t)$ in Vandermonde Matrix
 A: Perhaps the following line of thinking is clearer.
Suppose you pick some numbers $a_{ij}$ for $i \le j$ and form the matrix
$$U := \begin{bmatrix}a_{11} & {a_{12}} & {\cdots} & {a_{1 n}} \\ {0} & {a_{22}} & {\cdots} & {a_{2 n}} \\ {\vdots} & {\ddots} & {\ddots} & {\vdots} \\ {0} & {\cdots} & {0} & {a_{nn}}\end{bmatrix}.$$
Then you can show (by directly doing the matrix multiplication) that
$$VU = \begin{bmatrix}{f_{1}\left(x_{1}\right)} & {f_2(x_1)} & {\cdots} & {f_n(x_1)} \\ {f_{1}\left(x_{2}\right)} & {f_{2}\left(x_{2}\right)} & {\cdots} & {f_n(x_2)} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {f_{1}\left(x_{n}\right)} & {f_{2}\left(x_{n}\right)} & {\cdots} & {f_{n}\left(x_{n}\right)}\end{bmatrix}$$
where $f_j(t) := \sum_{i=1}^j a_{ij} t^{i-1}$.

Now, what if we choose the $a_{ij}$ in a particular way? Specifically, choose them to satisfy $\sum_{i=1}^j a_{ij} t^{i-1} = \prod_{i=1}^{j-1} (t-x_i)$ for $j \ge 2$ and $a_{11} = 1$.
Is this even possible? Well, $\prod_{i=1}^{j-1} (t-x_i)$ is a polynomial of degree $j-1$, so we can choose $a_{1j}, a_{2j}, \ldots, a_{jj}$ to be the coefficients. No problem.
Further note that the leading coefficient (coefficient of $t^{j-1}$) is $a_{jj}=1$.
So, in this case the above $U$ matrix has $1$s on the diagonal, like in your post.
Finally, note that $f_j(x_i) = 0$ for $i < j$ by our particular choice of $a_{ij}$, so the product matrix is lower triangular, as it appears in your post.
