Show that $p_i$ is polynomial function in $Pol_n(\mathbb{R})$ and $\{ \Phi(p_1),...,\Phi(p_{n+1}) \}$ is basis of $\mathbb{R}^{n+1}$ Let $n \in \mathbb{N}, Pol_n(\mathbb{R})$ be the vector space of the polynomial functions with degree $\le n$.
Given real numbers $x_1 \lt x_2 \lt ... \lt x_{n+1}$, consider
$$\Phi: Pol_n(\mathbb{R}) \rightarrow \mathbb{R}^{n+1}, \space f \mapsto \begin{pmatrix} f(x_1)\\f(x_2)\\\vdots\\f(x_{n+1}) \end{pmatrix}$$
and the functions
$$p_i:\mathbb{R}\rightarrow\mathbb{R}, \space x \mapsto \prod_{j=1,\space j \ne i}^{n+1}(x-x_j)$$ for all $i \in \{1,...,n+1 \}$. The index traverses all $j \in \{1,...,n+1 \} \setminus \{i\}$.
Show that the functions $p_i$ are polynomial functions in $Pol_n(\mathbb{R})$ and the set $\{ \Phi(p_1),...,\Phi(p_{n+1}) \}$ is a basis of $\mathbb{R}^{n+1}$.
I don't really know how to start, can someone help? Thanks in advance.

Every $p_i$ is a product of $n$ degree $1$ polynomials, thus is of degree $n$.
On the other hand, note that every $p_i$ satisfies:
$$p_i(x_j)=0\text{ if }j\neq i\text{ and }p_i(x_i)\neq 0.$$
So the matrix
$$\varphi:=\left[\Phi(p_1), \dots, \Phi(p_{n+1})\right]$$
is an $(n+1)\times(n+1)$ diagonal matrix with all diagonal elements not zero, and hence has a non-zero determinant. This means the set of columns of $\varphi$, which is exactly
$$\left\{\Phi(p_1), \dots, \Phi(p_{n+1})\right\},$$
 must be linearly independent. If $n+1$ vectors in $\mathbb{R}^{n+1}$ are linearly independent, they must form a basis of $\mathbb{R}^{n+1}$.
