TITLE: Why are Vandermonde matrices invertible?
QUESTION [24 upvotes]: A Vandermonde-matrix is a matrix of this form:
$$\begin{pmatrix}
 x_0^0 & \cdots & x_0^n \\
\vdots & \ddots & \vdots \\
 x_n^0 & \cdots & x_n^n
\end{pmatrix} \in \mathbb{R}^{(n+1) \times (n+1)}$$.
condition ☀ : $\forall i, j\in \{0, \dots, n\}: i\neq j \Rightarrow x_i \neq x_j$
Why are Vandermonde-matrices with ☀ always invertible?
I have tried to find a short argument for that. I know some ways to show that in principle:

rank is equal to dimension
all lines / rows are linear independence
determinant is not zero
find inverse

According to proofwiki, the determinant is
$$\displaystyle V_n = \prod_{1 \le i < j \le n} \left({x_j - x_i}\right)$$
There are two proofs for this determinant, but I've wondered if there is a simpler way to show that such matrices are invertible.

REPLY [3 votes]: For any $n+1$ distinct numbers $x_0, \ldots, x_n \in \mathbb{R}$, let 
$V(x_0,\ldots,x_n)$ and $D(x_0,\ldots,x_n)$ be a Vandermonde matrix and its determinant:
$$V(x_0,\ldots,x_n) = \begin{pmatrix}
 x_0^0 & \cdots & x_0^n \\
\vdots & \ddots & \vdots \\
 x_n^0 & \cdots & x_n^n
\end{pmatrix}\quad\text{ and }\quad
D(x_0,\ldots,x_n) = \det V(x_0,\ldots,x_n)
$$
It is clear $D(x_0) = 1 \ne 0$. Let us assume $D(x_0,\ldots,x_{m}) \ne 0$ for some $m < n$
and consider what happens to $D(x_0,\ldots,x_{m+1})$.
Viewed as a function of its last argument $x_{m+1}$, $D(x_0,\ldots,x_m,x)$ is a polynomial in $x$ with degree at most $m+1$.
Since this polynomial vanishes at $m+1$ different values $x_0, \ldots, x_{m}$ already, it cannot vanish on any other $x$ (in particular at $x_{m+1}$). Otherwise, $D(x_0,\ldots,x_m,x)$ will be identically zero. 
We know that $D(x_0,\ldots,x_m,x)$ isn't the zero polynomial. The leading coefficient of $x^{m+1}$ in $D(x_0,\ldots,x_m,x)$ is proportional to $D(x_0,\ldots,x_m)$ which is non-zero by induction assumption. 
By induction, we can conclude $D(x_0,\ldots,x_n) \ne 0$ and hence $V(x_0,\ldots,x_n)$ is invertible.