 Solution

The monomials $1,x,\dots ,x^n$ are linearly independent and span, so $\boxed{{\dim }_{\mathbb{C}}V=n+1}$. Use the factorial monomial basis for polynomial differentiation $p_j(x)=x^j/j!$, $0\le j\le n$. Direct differentiation gives $e_i(p_j)={\delta }_{ij}$. HenceIndeed every $\varphi \in V$ has the Taylor expansion $\varphi ={\sum }_{j=0}^n{e}_j(\varphi )p_j$; 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 $D{p}_0=0$ and $D{p}_j=p_{j-1}$ for $j\ge 1$. With column vectors of coefficients, its matrix $A$ has entries $A_{ij}=1$ if $j=i+1$ and zero otherwise, for indices $0\le i,j\le n$:The dual map is $D^{\ast }(\ell )=\ell \circ D$, not a conjugate-linear adjoint. Thus $D^{\ast }{e}_i=e_{i+1}$ for $i<n$ and $D^{\ast }{e}_n=0$. Therefore $\boxed{[D^{\ast }]=[D{]}^T}$ in the corresponding dual basis. For $n=0$ both matrices are the one-by-one zero matrix.