The problem reads:
$\frac{d}{dx}x^n = nx^{n-1}$, applying the differential operator to a function
If we use the monomial basis $\vec e_n = x^n$ then a function $$\vec u(x) = \sum _i u_i\vec e_i$$ is identified with the column vector with components $u_i$. The derivative of this function is again a function, $$\frac{d}{dx}u(x) = \frac{d}{dx} \sum_iu_i\vec e_i = \sum iu_i \vec e_{i-1}$$ This new vector can be expressed as $$\sum_iv_i \vec e_i$$ The differential operator $\frac{d}{dx}$ takes the fucntion u to the function v. What is the matrix that takes the vector $\vec u$ to $\vec v$?
So
$$ \vec{v}(x)=\frac{\mathbb{d}}{\mathbb{d}x}\vec{u}(x) $$
$$ \sum_{j\geq 0}v_j\hat{e}_j=\sum_{j\geq 0}ju_j\hat{e}_{j-1} $$
$$ v_{j-1}=ju_j $$
$$ \vec{v}= \begin{pmatrix} v_0 \\ v_1 \\ v_2 \\ \vdots \end{pmatrix}= \begin{pmatrix} u_1 \\ 2 u_2 \\ 3 u_3 \\ \vdots \end{pmatrix}= \begin{pmatrix} 0 && 1 && 0 && 0 && 0 && \dots \\ 0 && 0 && 2 && 0 && 0 && \dots \\ 0 && 0 && 0 && 3 && 0 && \dots \\ \vdots && \vdots && \vdots && \vdots && \ddots && \vdots \end{pmatrix} \begin{pmatrix} u_0 \\ u_1 \\ u_2 \\ u_3 \\ \vdots \\ \end{pmatrix} $$
$$ D=\begin{pmatrix} 0 && 1 && 0 && 0 && 0 && \dots \\ 0 && 0 && 2 && 0 && 0 && \dots \\ 0 && 0 && 0 && 3 && 0 && \dots \\ \vdots && \vdots && \vdots && \vdots && \ddots && \vdots \end{pmatrix} $$
In component form $D_{m,k}=k$ if $m=k-1$ and $0$ otherwise.