= Solution
The <monomials> $1,x,\ldots,x^n$ are linearly independent and span, so $\boxed{\dim_{\mathbb C}V=n+1}$. Use the <factorial monomial basis for polynomial differentiation> $p_j(x)=x^j/j!$, $0\leq j\leq n$. Direct differentiation gives $e_i(p_j)=\delta_{ij}$. Hence
$$
\boxed{(e_0,e_1,\ldots,e_n)\text{ is a basis of }V^*,\qquad(p_0,p_1,\ldots,p_n)\text{ is its dual basis in }V.}
$$
Indeed every $\phi\in V$ has the Taylor expansion $\phi=\sum_{j=0}^ne_j(\phi)p_j$; evaluation on the $p_j$ proves independence of the functionals. Derivative orders include zero. If one uses a convention excluding zero from $\mathbb N$, $e_0$ must explicitly be included, since all positive-order derivatives annihilate constants and cannot span the <dual space>.
Differentiation satisfies $Dp_0=0$ and $Dp_j=p_{j-1}$ for $j\geq1$. With column vectors of coefficients, its <matrix> $A$ has entries $A_{ij}=1$ if $j=i+1$ and zero otherwise, for indices $0\leq i,j\leq n$:
$$
\boxed{[D]=\begin{pmatrix}0&1&0&\cdots&0\\0&0&1&\cdots&0\\\vdots&\vdots&\ddots&\ddots&\vdots\\0&0&\cdots&0&1\\0&0&\cdots&0&0\end{pmatrix}.}
$$
The <dual map> is $D^*(\ell)=\ell\circ D$, not a conjugate-linear adjoint. Thus $D^*e_i=e_{i+1}$ for $i<n$ and $D^*e_n=0$. Therefore $\boxed{[D^*]=[D]^T}$ in the corresponding <dual basis>. For $n=0$ both matrices are the one-by-one zero <matrix>.
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