For a set of $n+1$ support points $(x_i, f_i)$, show that a polynomial $P(x;a_0,\dots, a_n)$ that satisfies $P(x_i)=f_i$ is unique.
If $P(x_i)=f_i$, then
$P(x_i) = \sum_{j=0}^n a_j x_i^j = f_i$ (1)
This linear equation system can be written in the matrix notation
$V \cdot \alpha = f$, (2)
$\alpha = (a_0, a_1, \dots, a_n)$. From there,
$\alpha = V^{-1} \cdot f$, if $V$ is invertible. We choose two polynomials $P_1(x), P_2(x)$ that satisfy (1), then
$\alpha_1 = V^{-1} \cdot f$
$\alpha_2 = V^{-1} \cdot f$
If we assume $P_2(x) \ne P_1(x)$, then
$\alpha_1 \ne \alpha_2$,
$V^{-1} \cdot f \ne V^{-1} \cdot f$
which is not true. This contradicts $P_2(x) \ne P_1(x)$, therefore $P_1(x) = P_2(x)=P(x)$.
The system
$$\sum_{k=0}^n p_kx_i^k=P(x_i),$$ for $i=0,\cdots n$ is linear in the $a_k$ and its matrix is of the Vandermonde form. As is well known, the determinant of a Vandermonde matrix is
$$\Delta=\prod_{0\le i<j\le n}(x_i-x_j),$$ which is nonzero when all $x$'s are different. Hence the solution of the system is unique.