Polynomial of degree $n$ with $n+1$ zeros 

Prove that if $P $ is a polynomial of degree $n$ with $n+1$ zeros $P$ must be zero


This can be proven easily by the fundamental Theorem of Algebra. However, how would one prove the statement above without using the fundamental Theorem of Algebra.
 A: Let $P(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$. W.L.O.G, let $x_1, x_2, \cdots, x_{n+1}$ be distinct values that are the solutions of $P(x) = 0$. We have
$$
\color{red}{
\begin{bmatrix}
1 & x_1 & x_1^2 & \cdots & x_1^n \\
1 & x_2 & x_2^2 & \cdots & x_2^n \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
1 & x_{n+1} & x_{n+1}^2 & \cdots & x_{n+1}^n 
\end{bmatrix}} \cdot
\begin{bmatrix}
a_0 \\
a_1 \\
\cdots \\
a_n
\end{bmatrix} =
\begin{bmatrix}
0 \\
0 \\
\cdots \\
0
\end{bmatrix}
$$
The square matrix in red is called a Vandermonde matrix and is invertible. It turns out from linear algebra that 
$$
\begin{bmatrix}
a_0 \\
a_1 \\
\cdots \\
a_n
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
\cdots \\
0
\end{bmatrix}
$$
A: You can easily prove, that if $a$ is a root of a polynomial $p$ the $p(x)=(x-a)q(x)$. Thus, inductively we have $p(x)=(x-a_1)\cdot...\cdot(x-a_{n+1})q(x)$. Since the degree of a product is the sum of the degrees, we obtain $n=deg(p)=n+1+deg(q)$. But only possibility for this to hold is $q\equiv p\equiv 0$.
