Zero Polynomials: Help Me Get out of a Circular Argument The following exercise appears at the beginning of Apostol's Calculus (Vol. 1), before any calculus has been developed, so it's not looking for a solution involving derivatives or Taylor's theorem:

Let $f(x)$ be a polynomial of degree $n$. If $f(x) = 0$ for $n+1$ distinct real values of $x$, then every
  coefficient $c_k$ is zero and $f(x)=0$ for all real $x$.

I can easily prove the second part - that is, $f(x)=0$ everywhere - by applying the factor theorem $n$ times, but this doesn't say anything about the coefficients. So I tried using an inductive argument, but I couldn't make it work, for two reasons:


*

*If $f(x)$ has degree $n+1$ and is zero everywhere, then $f(x) = xg(x)$ where $g(x)$ has degree $n$. We can say that $g(x)$ is zero for all $x\neq 0$, but not for $x=0$. Therefore we can't apply the induction step.

*Even if we got past problem 1., all we'd have shown is that $f(x)$ is equal to a polynomial of degree $n+1$ with all-zero coefficients. The argument is still not complete, because we don't know that every polynomial has a unique representation. The only way I know how to prove this uses the fact that if a polynomial is zero everywhere, it has all-zero coefficients - which is what we're trying to prove in the first place.
I found a proof that an everywhere-zero polynomial has all-zero coefficients, but this can't possibly be what Apostol is asking for from a beginning calculus student. Is there another, simpler proof?

If you know linear algebra, you can use Vandermonde determinant.
Let $f(x)=a_0+a_1x+\cdots+a_nx^n$, you want to solve for $a_0, a_1, \cdots, a_n$, knowing that $f$ has $n+1$ roots: $$f(x_0)=0, f(x_1)=0, \cdots, f(x_n)=0$$
Then we get $$
\begin{bmatrix}
1&x_0&\cdots&x_0^n\\
1&x_1&\cdots&x_1^n\\
\cdots &\cdots &\cdots &\cdots \\
1&x_n&\cdots&x_n^n
\end{bmatrix}
\begin{bmatrix}
a_0\\
a_1\\
\cdots\\
a_n\end{bmatrix} = 
\begin{bmatrix}0\\0\\\cdots\\0\end{bmatrix}
$$
This has nontrivial solution iff the determinant of the matrix on the left has determinant $0$. But that's the Vandermonde determinant! So this has nontrivial solution iff
$$\prod_{0\leq i<j\leq n}(x_j-x_i)=0$$
iff $x_i=x_j$ for some $0\leq i<j\leq n$, that is, two of the $n+1$ roots are equal. Since the $n+1$ roots are distinct, we have $a_0, \cdots, a_n = 0$, so $f(x)=0$
