give restriction on $n$-th degree polynomial $p(x)$ such that its derivatives form a basis of a polynomial vector space. So we have some $p\in V$ a $n$-th degree polynomial in a polynomial vector space $V$, question is when $\{p, p',p'',\ldots,p^{(n)}\}$ forms its basis.
So I want this set to i) span $V $ and be ii) linearly independent, first will be satisfied if for $p(x)=\sum^n_{i=0} b_ix^i$  we have $b_n$ is not zero. 
Second bit is more tricky since and here I need your help.
ii) will be satified if for $\sum_{i=0}^n a_i p^{(i)}(x)=\sum_{i=0}^n a_i \sum_{j=i}^n b_j \frac{j!}{(j-i)!} x^{j-i}=0 \Rightarrow a_i=0  \ \forall i=0,...,n$
but what next?
 A: I'm assuming that $V$ is the vector space of polynomials of degree $\le n$, so that $\dim V = n+1$.
If $A : V \to V$ is a linear map such that $A^n\ne 0$, $A^{n+1} = 0$ then the set $\{v, Av, \ldots, A^nv\}$ is a basis if and only if $v \notin \ker A^n$.
Namely, if  $\{v, Av, \ldots, A^nv\}$ is a basis then certainly $A^nv \ne 0$. Conversely, assume $v \notin\ker A^n$. To show that $\{v, Av, \ldots, A^nv\}$ it suffices to show linear independence. Assume $\sum_{i=0}^n \alpha_iA^iv = 0$ for some scalars $\alpha_i$.
Acting on the above equality with $A^n$ gives $0 = \sum_{i=0}^n \alpha_iA^{n+i}v = \alpha_0A^nv$ which implies $\alpha_0 = 0$ because $A^nv \ne 0$. Hence $0 = \sum_{i=1}^n \alpha_iA^iv$. Acting on this with $A^{n-1}$ gives $0 = \sum_{i=1}^n \alpha_iA^{n+i-1}v = \alpha_1 A^nv$ so $\alpha_1 = 0$. Continue inductively and conclude $\alpha_i = 0, \forall i = 0, \ldots,n$.
Now let $A$ be the derivative map given by $Ap = p'$. Clearly $A^{n+1} = 0$ and $A^n \ne 0$ because $A^n(x^n) = n! \ne 0$.
Hence $\{p, p', \ldots, p^{(n)}\} = \{p, Ap, \ldots, A^np\}$ is a basis for $V$ if and only if $A^np \ne 0$, which is true if and only if $\deg p = n$.
