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?

Given a polynomial $P$ define $(\Delta P)(x)=P(x+1)-P(x)$. Given a polynomial $P$ where $P(x)= a_0 + a_1x + \cdots + a_nx^n$, we claim that
$$
\Delta^{n}(P)(x)=n!a_{n}\tag{0}\label0.
$$
To see this write
$$
 P(x) = \sum_{k=0}^{n} a_k x^k=  \sum_{k=0}^{n} a_k \sum_{m=0}^{k}S(k,m)x^{\underline{m}}\tag{1}\label1
$$
where $S(k,m)$ is the stirling number of the second kind. and $x^{\underline{m}}$ is the falling factorial. Since $\Delta(x^{\underline m})=mx^{\underline{m-1}}$, $\Delta^{n}(x^{\underline {m}})=m^{\underline n}x^{\underline{m-n}}$. If $m<n$, then $\Delta^{n}(x^{\underline {m}})=0$. Hence taking $\Delta ^{n}$ of both sides of (1) yields that
$$
(\Delta^nP)(x)=n!S(n,n)a_{n}=n!a_{n}\tag{2}\label2.
$$
Now onto the problem.


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 $k$


We prove the claim by induction on $n$. The base case where $n=0$ yields that $a_0=0$ as desired. Suppose that the claim holds for all polynomials with degree less than $n$. Let $p$ be a polynomial where
$$
0=p(x)= a_0 + a_1x + \cdots + a_nx^n\tag{3}\label3
$$
for all $x\in\mathbb{R}$. Take $\Delta^{n}$ of both sides of equation \eqref{3} and use equation \eqref{0}, to get that $a_{n}=0$. The induction hypothesis now implies the result.
