Solving for the Nth divergence in any coordinate system

Ok, so I was rereading through your posts 19, 21, 23, and 25, and I think I understand them better now. I also think I MIGHT have found what I was looking for $$\nabla^{n} f = \sum_{i=1}^{m} \left [ \partial_{x_{i}}^{n} \left [ f \right ] \right ] = \sum_{i=1}^{m} \left [ \left [ \partial_{x_{i}} \right ]^{n} \left [ f \right ] \right ] = \sum_{i=1}^{m} \left [ \left [ \frac {\partial } {\partial {x_{i}}} \right ] ^{n} \left [ f \right ] \right ] = \sum_{i=1}^{m_{x}} \left ( \left [ \sum_{j=1}^{m_{q}} \left [ \frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j} \right ] \right ]^{n} \left [ f \right ] \right )$$ and the ##\nabla^{n}## operator itself $$\nabla^{n}= \sum_{i=1}^{m_{x}} \left ( \left [ \sum_{j=1}^{m_{q}} \left [ \frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j} \right ] \right ]^{n} \right )$$ where ##x## represents the Cartesian Basis, and ##q## represents the arbitrary Basis. There are ##m_x## dimensions in the Cartesian Basis and ##m_q## dimensions in the arbitrary Basis.
I think this holds, since, in Einstein Notation, we have the multivariable chain rule as $$\frac{\partial}{\partial x^i}=\frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j}$$

Is the like meant to indicate that that is correct? :D

I observe my #25 and your #36 coincide.

Also, would it be necessary for ##m_x=m_q## for that equation $$\nabla^{n}= \sum_{i=1}^{m_{x}} \left ( \left [ \sum_{j=1}^{m_{q}} \left [ \frac{\partial q^j}{\partial x^i}\frac{\partial }{\partial q^j} \right ] \right ]^{n} \right )$$ to work?

We need same number of parameters to describe same space.

What do I do with mixed terms (I.e., more than 1 basis vector involved), such as $$-2\cos \phi\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}$$ in my calculation? Do I just discard those?

Vanilla Gorilla said:

What do I do with mixed terms (I.e., more than 1 basis vector involved), such as −2cos⁡ϕ∂∂rsin⁡ϕr∂∂ϕ in my calculation? Do I just discard those?

[tex]-2\cos \phi\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}[/tex]
We have no problem on the leftest
[tex]-2\cos \phi[/tex]. Applying produclt rule of differentiaion,
[tex]\frac{\partial}{\partial r}\frac{\sin \phi}{r}\frac{\partial}{\partial \phi}[/tex]
[tex]=[\frac{\partial}{\partial r}\frac{\sin \phi}{r}]\frac{\partial}{\partial \phi}+\frac{\sin \phi}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \phi}[/tex]
[tex]=-\frac{\sin \phi}{r^2}\frac{\partial}{\partial \phi}+\frac{\sin \phi}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \phi}[/tex]

As for change of order of operators for an example
[tex]\frac{d}{dx}xA=A+x\frac{d}{dx}A[/tex]
As operator we may delete A as
[tex]\frac{d}{dx}x=1+x\frac{d}{dx}[/tex]
[tex]1=\frac{d}{dx}x-x\frac{d}{dx}=[\frac{d}{dx},x][/tex]
In general
[tex][\frac{d}{dx},f(x)]=f'(x)[/tex]