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?
 A: When going to $k+1$, one begins doing some algebraic "hokus pokus" trying hard to get some expressions where to use the inductive hypothesis, which is the leading light almost anyone doing induction follows: how, where, when do I "put in" the inductive hypothesis to use that and get what I want.
$\mbox{-}$ DonAntonio (from comments just placed as answer)
A: 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.)
$$
\begin{align}
(a-1)(a^{n-1} + a^{n-2} + \dots + a + 1) &= a(a^{n-1} + a^{n-2} + \dots + a + 1) - (a^{n-1} + a^{n-2} + \dots + a + 1)\\
&= a^n + a^{n-1} + \dots + a^2 + a - (a^{n-1} + a^{n-2} + \dots + a + 1) \\
&= a^n - 1.
\end{align}
$$
A: Observe that $$a^{n+1}-1=(a^n-1)+(a^{n+1}-a^n)=(a^n-1)+(a-1)a^n.$$ So if $a^n-1=(a-1)(1+...+a^{n-1})$ then $$a^{n+1}-1=(a^n-1)+(a-1)a^n=$$ $$=(a-1)(1+...+a^{n-1})+(a-1)a^n=$$ $$=(a-1)([1+...+a^{n-1}]+a^n)=$$ $$=(a-1)(1+...+a^n).$$ This is a frequently used method. For example to prove that $1+...+n=F(n)=n(n+1)/2$ for all $n\in \mathbb N,$ after noting that it is valid when $n=1,$ we have $1+...+(n+1)=(1+...+n)+(n+1).$ 
So if $(1+...+n)=F(n)$ then $1+...+(n+1)=(1+...+n)+(n+1)=F(n)+(n+1)=F(n+1). $ 
A: If you have any inclination towards mathematics at all, you know that the equation
$a^n - 1 = 0$
can be solved by letting $a = 1$. It was not that long ago that you 'could just see' that
$x^2 - 1 = (x - 1) (x + 1)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ (1)
When you divide $x^3 - 1$ by $x -1$, you realize you are on the road to discovery when get
$x^3 - 1 = (x -1) (x^2 + x^1 + x^0)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,$ (2)
Emboldened by all this, you want to prove that
$x^6 - 1 = (x -1) (x^5 + x^4 + x^3 + x^2 + x^1 + x^0)\;\;\;\;\;$ (3)
by making use of (1) and (2):
$x^6 - 1 = (x^3 + 1) (x^3 - 1) = (x^3 + 1) (x - 1) (x^2 + x^1 + x^0) =$ 
$\;\;\;(x - 1) \,(x^3 + 1)\, (x^2 + x^1 + x^0)=$
$\;\;\;(x - 1)\,[ \,x^3\, (x^2 + x^1 + x^0)\,+ (x^2 + x^1 + x^0)\,]$
You don't even bother finishing the work - you 'can see it'.
Before getting formal and using strong induction (also called the second principle of finite induction), you have a nagging feeling (odd exponents), so you write down
$x^7 - 1 = x (x^3 + 1) (x^3 - 1) + (x -1)$
Amazingly, you don't have to go on. Working on the right side, you factor out $x - 1$ and can 'see' that the coefficients are all '+1' and that each $x^k \;\;\; \text {for } 0 \le k \le 6$ appears.
Suddenly you hear laughter! A friend explains to you that they can see the same thing for the general case without doing any work! They tell you to take a look at the article Telescoping Sum. As they walk away you hear them say something about a geometric series.
Now you realize that you can figure all this out by using the powerful capital-sigma notation:
${\displaystyle \sum _{i{\mathop {=}}m}^{n}a_{i}=a_{m}+a_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_{n}}$
If you just accept that you can do so many general manipulations under this notation, you can simnply dispense with induction if you still feel it is necessary to prove this.
