How does this differentiation map work? This example is from Linear Algebra Done Right by Sheldon Axler. I don't quite get it.

Example: Suppose $D \in \mathcal{L}\Big(\mathcal{P}_3(\mathbf{R}, \mathcal{P}(\mathbf{R}))\Big)$ is the differentiation map defined by $Dp = p'$. Find the matrix of $D$ with respect to the standard bases of $\mathcal{P}_3(\mathbf{R})$ and $\mathcal{P}_2(\mathbf{R})$.
Solution: Because $(x^n)' = nx^{n-1}$, the matrix of $T$ with respect to the standard bases is the $3$-by-$4$ matrix below: $$\mathcal{M}(D) = \begin{pmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3\end{pmatrix}.$$

 A: Since differentiation is a linear map from polynomials of degree 3 to polynomials of degree 2, we can express it as matrix multiplication:
$$p_3(x)=a_0+a_1x+a_2x^2+a_3x^3\to p_3'(x)=a_1+2a_2x+3a_3x^2$$
$$\begin{bmatrix} 
{D}_{11} & D_{12} & D_{13} & D_{14}\\
D_{21} & D_{22} & D_{23} & D_{24}\\
D_{31} & D_{32} & D_{33} & D_{34}\end{bmatrix}
%\bigg[
\begin{bmatrix} 
{a}_0\\a_1\\a_2\\a_3
\end{bmatrix}
%\bigg]
=
\begin{bmatrix}
a_1\\
2a_2\\
3a_3
\end{bmatrix}
$$
You can then find the $D_{ij}$ that determine the matrix.
A: In Axler's example, $p$ is identified with the vector whose $k$th entry is the $x^k$ coefficient in $p(x)$, hereafter $[x^k]p(x)$. Then $D$ can be identified with the matrix whose ${}_{ij}$ entries $D_{ij}$ satisfy$$\sum_jD_{ij}p_j=p^\prime_i=[x^i]p^\prime=(i+1)[x^{i+1}]p=(i+1)p_{i+1}=\sum_jj\delta_{i+1,\,j}p_j,$$i.e. $D_{ij}=j\delta_{i+1,\,j}$, where for convenience I've somewhat unconventionally started matrix indices at monomials' minimum degree $0$ rather than $1$. If we start indices at $1$,  the result is$$D_{ij}=(j-1)\delta_{i+1,\,j}=i\delta_{i+1,\,j}.$$
