Help in proving that $\nabla\cdot (r^n \hat r)=(n+2)r^{n-1}$ 
Show that$$\nabla \cdot  (r^n \hat r)=(n+2)r^{n-1}$$  where $\hat r$ is the unit vector along $\bar r$.

Please give me some hint. I am clueless as of now.

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\begin{align}
\color{#0000ff}{\large\nabla\cdot\pars{r^{n}\,\hat{r}}}
&=
\nabla\cdot\pars{r^{n - 1}\vec{r}}
=
\nabla\pars{r^{n - 1}}\cdot\vec{r} + r^{n - 1}\nabla\cdot\vec{r}
=
\bracks{\totald{r^{n - 1}}{r}\,{\vec{r} \over r}}\cdot\vec{r}
+ r^{n - 1}\times 3
\\[3mm]&=
\pars{n - 1}r^{n - 2}\,{r^{2} \over r} + 3r^{n - 1}
=
\color{#0000ff}{\large\pars{n + 2}r^{n - 1}}
\end{align}

We have used the identities:
$$
\nabla\cdot\pars{\varphi\vec{A}}
= \nabla\varphi\cdot\vec{A} + \varphi\nabla\cdot\vec{A}\,,\qquad
\nabla\fermi\pars{r} = \totald{\fermi\pars{r}}{r}\,{\vec{r} \over r}\,,\qquad
\nabla\cdot\vec{r} = 3
$$

