How do I show that $6(4^n-1)$ is a multiple of $9$ for all $n\in \mathbb{N}$? How do I show that $6(4^n-1)$ is a multiple of $9$ for all $n\in \mathbb{N}$? I'm not so keen on divisibility tricks. Any help is appreciated.

The following identity is often useful:
$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1).$$
Like many identities, it is easy to verify: just expand the right-hand side, and note that almost everything cancels. 
If we put $x=4$, then it follows from the above identity that $4-1$ divides $4^n-1$.
Remark: For $x\ne 1$, the identity can be rewritten as 
$$1+x+x^2+\cdots +x^{n-1}=\frac{x^n -1}{x-1}.$$ 
This is one version of the formula for the sum of a finite geometric series. Everything is connected to everything else.
