Proof that monomials of $\mathbb{P}_n(x)$ are linearly independent My Professor's proof is based on this implication: if
$\forall x \in \mathbb{R},p(x) = c_0+c_1x+c_2x^2+...+c_nx^n=0$
, then
$p(0) = 0$ $\wedge$ $p'(0)=0$ $\wedge$ $p''(0)=0$ $\wedge$ $...$ $\wedge$ $p^n(0)=0$. Why is this implication true?
 A: The idea is that if $f(x) = 0$ for all $x \in \mathbb R$, then $f'(x) = 0$ for all $x \in \mathbb R$. We can prove this by definition of a derivative: for any $x \in \mathbb R$,
$$
   f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{0-0}{h} = \lim_{h \to 0} 0 = 0.
$$
Repeating this argument tell us that if $f(x) = 0$ for all $x \in \mathbb R$, then $f'(x), f''(x), \dots$ are all identically $0$, so in particular $f(0), f'(0), f''(0), \dots$ are all $0$.
This works for any function, but in particular it works for any polynomial.
A: If you are just trying to prove that $1,x,\dots,x^n$ is linearly independent. The following proof might help.
Proof: Suppose that $1,x,\dots,x^n$ is linearly dependent. Then there exists a polynomial $x^j$ such that $x^j\in span(1,\dots,x^{j-1})$. Thus there exist scalars $a_0,\dots,a_{j-1}$ such that $$x^j=a_01+\dots+a_{j-1}x^{j-1}$$
Differentiating both sides $j$ times we get a contradiction as the L.H.S is non-zero and the R.H.S is zero.
Thus the list $1,x,\dots,x^n$ is linearly independent. Then any linear combination $a_01+a_1x+\dots +a_nx^n=0$ if and only if $a_0=a_1=\cdots =a_n=0$.
