Characteristic of a field and basis of a dual space. Let $K$ be a field. $E=K_n[x]=\{p(x)\in K[x]|\text{degree}(p(x))\leq n\}$

For a $p(x)\in E$ we define $p^{(0)}(x)=p(x)$ and for every positive integer $i$, $p^{(i)}(x)=(p^{(i-1)})'(x)$. For $i\in \{0,...,n\}$ consider $w_i:E\longrightarrow K$ given by $w_i(p(x))=p^{(i)}(1)$. Prove that $B=(w_0,...,w_n)$ is basis of $E^*$ if and only if $K$ has characteristic $0$ or characteristic $p&gt;n$.

I can't solve this problem because I don't even understand the problem statement.

First of all, what is $p^{(i)}(x)$? Of course I understand that $p(x)=a_0+a_1x+...+a_nx^n$ so $p^{(3)}(x)$ would be $((a_0+a_1x+...+a_nx^n)^2)'$?

Second of all, what is the characteristic of $K$?

Thank you for your time and excuse me if I made any grammar mistake, my english is not too good.
$p^{(3)}$ would be $((a_0+a_1x+...+a_nx^n)^2)'$.

No, you have to build it up so

$$
(a_0+a_1x+...+a_nx^n)^{(3)}=
(a_0+a_1x+...+a_nx^n)^{(2)}{'}=
(a_0+a_1x+...+a_nx^n)^{(1)}{''}=
(a_0+a_1x+...+a_nx^n)^{(0)}{'''}=
(a_0+a_1x+...+a_nx^n){'''}
$$
and you now just have to differentiate the polynomial three times.

The characteristic of a field $\mathbb{F}$ is just the least positive $k$ such that $k\cdot 1=0$; or if there are none, $0$. In fact such a $k$ will be a prime number $p$. 

That answers your questions, good luck with the calculation to show the $w_i$ are linearly independent exactly when the characteristic is $0$ or greater than $n$.
You can take derivatives formally. 

Every finite field with $p^n$ elements has characteristic $p$ which is the smallest nonzero integer $k$ such that $1+1+\cdots+1=k\times 1=0.$ So you'd have modulo $k$ operations.

$R$ and $C$ have characteristic zero, sums of the additive identity are never zero.