$p$-polynomial of $n$'th degree, $q(x)=p[x,x_1,x_2,...,x_k]$, prove that q has the same leading coefficient.

Question

So I have a polynomial $p$ of $n$'th degree and q given by $q(x)=p[x,x_1,x_2,...,x_k]$, meaning that for $x$ it gives back the leading coefficient in interpolation of $p$ on points $x,x_1,...,x_k$. Prove that q is of degree $n-k$, and the coefficient beside $x^{n-k}$ is the same, as the one beside $x^n$ in $p$.

Answer 1

For the original question, a counterexample. $f(x)=x^2$ interpolated at two points is linear, $Ax+B$, and $A$ is dependent on the two points, not only the coefficient of $x^2$ in $f$.

The statement in the new question is correct, because if you allow the interpolation points to include $x$, and substitute $x$ into the result, you recover the original polynomial completely, not only its highest degree term, and $c(x) \prod (x -x_i)$ is the unique part of this expression with highest degree of $x$.