<s>[INST] Lack of understanding of steps for second principle of finite induction for: $a^n -1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1)$ The problem i'm trying to solve is:

Use the second principle of finite induction to establish that:
  $$a^n -1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1)$$
  for all $n \ge 1$

I have found a solution online that reads:
$$ \mbox{For }k=1: \ a^1-1 =a-1=(a-1)(a^0)=a-1 $$
\begin{align}
k \Rightarrow k+1: \ a^{k+1} -1 &= a^{k+1}-a^k-a+a^k+a-1\\
&=a(a^k-1)+a^k-1-a(a^{k-1} -1)\\
&=(a+1)(a^k-1)-a(a^{k-1}-1)
\end{align}
Use second principle of induction for $k,\ k-1$:
\begin{align}
&=(a+1)[(a-1)(a^{k-1}+a^{k-2}+\cdots+a+1)] \\
&\qquad \;  \; -a[(a-1)(a^{k-2}+a^{k-3}+\cdots+a+1)]\\
&=(a-1)[(a)(a^{k-1}+a^{k-2}+\cdots+a+1)\\
&\qquad \quad \ \, +(1)(a^{k-1}+a^{k-2}+\cdots+a+1)\\
&\qquad \quad \ \,-(a)(a^{k-2}+a^{k-3}+\cdots+a+1)]\\
&=(a-1)[(a^k+a^{k-1}+a^{k-2}+\cdots+a^2+a)+1]
\end{align}
$\therefore$ works for $k+1$
My problem:
I understand all the steps used for the principle of second induction however i do not understand how you are supposed to deduce that you need:
$$a^{k+1} -1 = (a+1)(a^k-1)-a(a^{k-1}-1)$$
To clarify i understand the logic that leads to the equation above, but do not understand how you know to try to make it. Is there something specific about the format $ (a+1)(a^k-1)-a(a^{k-1}-1)$ that allows you to know it is needed, or has the format just been found through any possible means that leads to $(a^k-1)$ and $(a^{k-1}-1)$ being used?
 [/INST] I'm just going to do the induction step here, for $k \to k + 1$, to show that it can be done in a simpler and more transparent way.
So assume you already know that 
$$a^k - 1 = a^{k-1} + a^{k-2} + \dots + a + 1.$$
Then, using the induction hypothesis on the second line, we have
\begin{align}
(a-1)(a^k + a^{k-1} + \dots + a + 1) &= (a-1)a^k + (a-1)(a^{k-1} + a^{k-2} + \dots + a + 1) \\
&= a^{k+1} - a^k + a^k - 1 \\
&= a^{k+1} - 1.
\end{align}
This proves the result for $k + 1$.
Edit Just in case you haven't seen it, here is the proof without induction. (Technically, there's probably some kind of induction hidden somewhere inside it, but it's not formalized.)
$$
\begin{align}
(a-1)(a^{n-1} + a^{n-2} + \dots + a + 1) &= a(a^{n-1} + a^{n-2} + \dots + a + 1) - (a^{n-1} + a^{n-2} + \dots + a + 1)\\
&= a^n + a^{n-1} + \dots + a^2 + a - (a^{n-1} + a^{n-2} + \dots + a + 1) \\
&= a^n - 1.
\end{align}
$$
</s>