Say we have a vector field $\vec{a}$ such that
$$\vec{a}=\begin{bmatrix}a_1(x_1,x_2,\ldots,x_n) \\ a_2(x_1,x_2,\ldots,x_n)\\ \vdots \\a_n(x_1,x_2,\ldots,x_n)\end{bmatrix}.$$
Then in an orthonormal basis the divergence $\vec{a}\cdot\vec{\nabla}$ is $$\vec{a}\cdot\vec{\nabla}=\frac{\partial[a_j(x_k)]}{\partial x_i}\vec{e_i}\cdot\vec{e_j}=\frac{\partial[a_j(x_k)]}{\partial x_i}\delta_{ij}=\frac{\partial[a_j(x_k)]}{\partial x_j}\left(=\frac{\partial a_j}{\partial x_k}\frac{\partial x_k}{\partial x_j}=\frac{\partial a_j}{\partial x_k}\delta_{kj}=\frac{\partial a_j}{\partial x_j}\right).$$
Einstein's summation convention has been applied here, therefore
$$\vec{a}\cdot\vec{\nabla}=\sum_{j=1}^n\frac{\partial a_j}{\partial x_j}.$$Let us also have a position vector denoted by $\vec{p}$ whose coordinates in the same orthonormal basis are given as $$\vec{p}=\begin{bmatrix}x_1 \\ x_2\\ \vdots \\ x_n\end{bmatrix}.$$
Now scaling the position vector by the divergence yields a new vector:
$$(\vec{a}\cdot\vec{\nabla})\vec{p}=\begin{bmatrix}x_1\frac{\partial a_j}{\partial x_j} \\ x_2\frac{\partial a_j}{\partial x_j}\\ \vdots \\ x_n \frac{\partial a_j}{\partial x_j}\end{bmatrix}.$$
Answer key shows $$(\vec{a}\cdot\vec{\nabla})\vec{p}=\vec{a},$$ or, in other words,
$$\begin{bmatrix}x_1\frac{\partial a_j}{\partial x_j} \\ x_2\frac{\partial a_j}{\partial x_j}\\ \vdots \\x_n\frac{\partial a_j}{\partial x_j} \end{bmatrix} = \begin{bmatrix} a_1(x_1,x_2,\ldots,x_n) \\ a_2(x_1,x_2,\ldots,x_n)\\ \vdots \\a_n(x_1,x_2,\ldots,x_n)\end{bmatrix}.$$
How?
