Prove by induction that for $a\ge1$, $a^{n}-1 = (a-1)(1+a+a^{2}+...+a^{n-1})$, for $n\ge1$, $n\in\mathbb{Z^+}$.

Question

I have doubts about this problem about induction. The statement says:

Prove by induction that for $a\ge 1 \Rightarrow$ $$ a^n-1 = (a-1)\left(1 + a + a^2 + \ldots + a^{n-1}\right). $$ for $n\in\mathbb{Z^+}$

I can see that it is valid for $a=1$ and $n = 1$, but I cannot go on from there.

The associated set is $A = \{2, 3, 4, \ldots \}$, but and the proposition, it's all that or just $a^n -1 = (a-1)\left(1 + a + a^2 + \ldots + a^{n-1}\right)$ ?

Thank you!

Answer 1

Hint for the induction step:

$a^{n+1}-1=a^{n+1}-a^n+a^n-1=(a-1)a^n+(a-1)(1+a+a^2+\cdots+a^{n-1})$

Answer 2

Note that $$ a^{n+1}-1=a^{n+1}-a+a-1= a\cdot (a^n-1)+(a-1)$$