Regarding the solution of finding the remainder of $g(x^{12})$ divided by $g(x)$ 
Let $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1.$ What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$?

I'm reading the solution for this and I don't understand how can someone come up with it.
It states

We have that $g(x^{12}) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1$

which is still very clear. Then it goes to say that

Note that $(x - 1)g(x) = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1) = x^6 - 1$

but I am not "noting" this. Where is this expression coming form? There is nothing natural to it.
Furhtermore they state that

Also,
\begin{align*}
g(x^{12}) - 6 &= (x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1) - 6 \\
&= (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).
\end{align*}

which is also not very natural to come up with. Where is this $6$ coming from?
The rest follows from $g(x^{12})-6$ being a multiple of $g$ and the remainder is thusfore $6$. I hate reading these kinds of solutions there is nothing to be learnt from this. It's just a bunch of sentences with not substance to it.


$(x - 1)g(x) = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1) = x^6 - 1$

That product is "telescoping":
$$\begin{array}{ccccccc}
(x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1)
&= &x^6 &+ x^5 &+ x^4 &+ x^3 &+ x^2 &+ x \\
&  &    &- x^5 &- x^4 &- x^3 &- x^2 &- x &-1 \\
&=& x^6&&&&&&-1
\end{array}$$
More generally: $$\sum_{k=0}^n x^k = \frac{x^{n+1}-1}{x-1}$$
