Showing that an $n^{th} $ degree polynomial is identically null

Question

Let $f:\mathbb{C} \to \mathbb{C} $ be a polynomial of $n^{th} $ degree. If $\exists x_1,x_2,...x_{n+1}\in \mathbb{C} $ (they are all distinct numbers) so that $f(x_k) =0, \forall k=\overline{1,n+1}$,then $f(x) =0,\forall x\in \mathbb{C} $.
I think I should use the fact that an $n^{th} $ degree polynomial has $n$ roots(with different multiplicities), but I don't know why the $(n+1)^{th}$ root implies that the polynomial is identically null.

Answer 1

Hint: You can factor the polynomial as $f(x) = A(x-x_0)\cdots(x-x_n)$. $x_{n+1}-x_j \neq 0$ for any $1\le j\le n$, so what does $A$ have to be?