Zero function implies zero polynomial. I'm trying to help someone with a problem in Apostol's book (Chapter 1 BTW, so before basically any calculus concepts are covered) at the moment and I'm stumped on a question.
I'm trying to prove that if $p$ is a polynomial of degree $n$, that is where
$$p(x) = a_0 + a_1x + \cdots + a_nx^n$$
for some real numbers $a_0, \dots, a_n$, and if $p(x) = 0$ for all $x\in \Bbb R$, then $a_k = 0$ for all $k$.
Looking through the site, I find this question, but the solution given uses the derivative.  But this before the definition of the derivative in Apostol's book, so I can't use that to prove this.  I also know that we can use linear algebra to solve this, but pretend I don't understand the concept of linear independence either as Apostol's book doesn't presuppose that.  Then what can we do to prove this?  It feels like there should be a proof by induction possible, but I'm not seeing how to do the induction step.
My Attempt: Proving that $a_0 = 0$ is trivial by evaluating $p(0)$.  But then I'm left with
$$p(x) = x(a_1 + \cdots +a_nx^{n-1})$$
Here I see that for all $x\ne 0$, $a_1 + \cdots a_nx^{n-1}=0$.  But because of that $x\ne 0$ part, that means I can't use the same trick to show that $a_1 = 0$.
Any ideas?

Note that according to the Fundamental Theorem of Algebra a polynomial of degree $n$ has exactly $n$ roots.
Now your function has infinitely many zeros, therefore it can not be a polynomial of degree $n$ for any $n$.
Thus all the coefficients are zero which makes your function to be identically zero.    
