Another way to prove that if n$^{th}$-degree polynomial $r(z)$ is zero at $n+1$ points in the plane, $r(z)\equiv 0$? The original problem is as follows

Let $p$ and $q$ be polynomials of degree $n$. If $p(z)=q(z)$ at $n+1$ distinct points of the plane, the $p(z)=q(z)$ for all $z\in \mathbb{C}$.

I attempted showing this is by taking the polynomial $r(z)=p(z)-q(z)$. Since both $p(z)$ and $q(z)$ are both polynomials of degree $n$, $r(z)$ can either be identically zero ($r(z)\equiv 0$), or it can be a polynomial of a degree less than $n$. However, if $p(z)=q(z)$ at $n+1$ distinct points, then $r(z)=0$ at $n+1$ points, which can only happen if $r(z)\equiv 0$ by the fundamental theorem of algebra. Hence, $p(z)=q(z)$. $\blacksquare$.
Is this the correct way to prove this? Is there another way? I'm asking this because I immediately jumped to this proof, and usually I'm wrong with these problems that are at the end of the problem set for a subsection.

When people say “Fundamental Theorem of Algebra”, they usually mean the result that every complex polynomial has a complex root. This is far simpler than that, since it’s true over any field at all, not even of characteristic zero.
For a proof, let $\rho_1,\dots,\rho_n,\rho_{n+1}$ be the roots. Then by Euclidean division, you show that $X-\rho_1$ divides your original $r(X)$, in such a way that the quotient polynomial still vanishes at the other $\rho_i$’s. Then you proceed by induction to show that $\prod_i(X-\rho_i)$ divides your original. But the product of all the $X-\rho_i$ is of degree $n+1$. So that original $r(X)$ has to have been zero to start with.
