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.

You can also use Cartesian coordinates and using the fact that $r \hat{r} = \vec{r} = (x,y,z)$.
\begin{align}
r^n \hat{r} &= r^{n-1} (x,y,z) \\
\nabla \cdot r^n \hat{r} & = \partial_x(r^{n-1}x) + \partial_y(r^{n-1}y) + \partial_z(r^{n-1}z) 
\end{align}
Each term can be calculated:
$\partial_x(r^{n-1}x) = r^{n-1} + x (n-1) r^{n-2} \partial_x r$
$\partial_x r = \frac{x}{r}$. (Here, I used the fact that $r = \sqrt{x^2+y^2+z^2}$.)
The terms involving $y$ and $z$ are exactly the same with $y$ and $z$ replacing $x$.
And so,
\begin{align}
\nabla \cdot r^n \hat{r}& = r^{n-1} + x(n-1)r^{n-2}\frac{x}{r} + r^{n-1} + y(n-1)r^{n-2}\frac{y}{r} + r^{n-1} + z(n-1)r^{n-2}\frac{z}{r}\\
& = 3r^{n-1}+(n-1)(x^2 + y^2 + z^2)r^{n-3} \\
& = 3r^{n-1} + (n-1)r^2 r^{n-3} \\
& = (n+2)r^{n-1}
\end{align}
Of course, polar is the easiest :)
