Polynomial equal to polynomial of lower degree I am studying Linear Algebra Done Right, chapter 2 problem 6 states:

Prove that the real vector space consisting of all continuous real valued functions on the interval $[0,1]$ is infinite dimensional.

My solution:
Consider the sequence of functions $x, x^2, x^3, \dots$ 
This is a linearly independent infinite sequence of functions so clearly this space cannot have a finite basis.
However this prove relies on the fact that no $x^n$ is a linear combination of the previous terms. In other words, is it possible for a polynomial of degree $n$ to be equal to a polynomial of degree less than $n$. I believe this is not possible, but does anyone know how to prove this? More specifically, could the following equation ever be true for all $x$?
$x^n = \sum\limits_{k=1}^{n-1} a_kx^k$ where each $a_k \in \mathbb R$

Then the polynomial $\displaystyle x^n-\sum_{k=0}^{n-1}a_kx^k$ would have infinitely many roots, but it can have $n$, at most.

Another way of dealing with this problem is based upon defining polynomials (in one variable $x$) as expressions of the type $a_0+a_1x+a_2x^2+\cdots+a_nx^n$, where $n\in\{0,1,2,\dots\}$ and each $a_n$ is real. Under this definition, the polynomial $a_0+a_1x+a_2x^2+\cdots+a_nx^n$ is equal to the polynomial $b_0+b_1x+a_2x^2+\cdots+b_nx^n$ if and only if the coefficients are equal, that is, if and only if $a_0=b_0$, $a_1=b_1$, and so on. Under this definition, the problem discussed here is trivial.
What did I prove above then? Well, for each $P(x)\in\mathbb{R}[x]$, there is a corresponding polynomial function from $\mathbb R$ into $\mathbb R$. What I proved above is that this correspondence is one-to-one — when we are dealing with $\mathbb R$. It is still one-to-one if we are dealing with any field with charactristic $0$, such as $\mathbb Q$ or $\mathbb C$. But this is not true in general. For instance, if our field is $\mathbb{F}_2$, then $x$ and $x^2$ are distinct polynomials. But they correspond to the same polynomial function.
