This is the solution according to my book.

> 
> Solution. For convenience, let us denote the polynomials as
>  $$p\_0=1,p\_1=x,p\_1=x^2,\dots,p\_n=x^n$$
>  We must show that the only coefficients satisfying the vector equation
>  \begin{equation}\tag{5}
> a\_0 p\_0 + a\_1 p\_1 + a\_2 p\_2 + \cdots + a\_n p\_n = 0
> \end{equation}
>  are
>  $$a\_0=a\_1=a\_2=\cdots=a\_n=0$$
>  But $(5)$ is equivalent to the statement that
>  \begin{equation}\tag{6}
> a\_0+a\_1 x+a\_2 x^2+\cdots+a\_n x^n=0
> \end{equation}
>  for all $x$ in $(-\infty,\infty)$, so we must show that this is true if and only if each coefficient in $(6)$ is zero. To see that this is so, recall from algebra that a nonzero polynomial of degree $n$ has at most $n$ distinct roots. That being the case, each coefficient in $(6)$ must be zero, for otherwise the left side of the equation would be a nonzero polynomial with infinitely many roots. Thus, $(5)$ has only the trivial solution.
> 
> 
> 

(original here: [http://i.stack.imgur.com/sp7vY.png](https://i.stack.imgur.com/sp7vY.png))

I don't really understand the conclusion that it makes. Why would the left side of the equation be a non-zero polynomial with infinitely many roots? Doesn't the sentence preceding it say that a non-zero polynomial of degree $n$ has at most $n$ distinct roots? Does that not contradict the statement that it would have infinitely many roots? And how does this all end up meaning that each coefficient must be zero?

Can somebody clarify this a bit for me?