<s>[INST] Verification for proof by strong induction of $a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + a^{n-3}+···+a + 1)$ I'm trying to prove by strong induction that
$$a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + a^{n-3}+···+a  + 1),$$
for $n \geq 1$.
By strong induction, I'd like to know if my solution is valid. What I did was:

*

*proof for $n = 1$
$$a - 1 = (a-1)(1)$$

*

*$n = k$
$$a^k - 1 = (a - 1)(a^{k-1} + a^{k-2} + a^{k-3}+···+a  + 1)$$


*proof for $n = k + 1$
We want to show that:
$$a^{k+1} - 1 = (a - 1)(a^k + a^{k-1} + a^{k-2} + a^{k-3}+···+a  + 1)$$
as we know, $$a^{k+1} - 1 = a^{k+1} - a^k + a^k - 1$$
so we rewrite the statement and use that fact that for $n = k$ the property holds.
\begin{eqnarray*}
a^{k+1} - 1 &=& a^{k+1} - a^k +a^k - 1\\ 
&=& a^{k+1} - a^k + (a - 1)(a^{k-1} + a^{k-2} + a^{k-3}+···+a  + 1),
\end{eqnarray*}
now factorizing $a^k$ we get:
$$a^{k+1} - 1 = a^k (a-1) + (a - 1)(a^{k-1} + a^{k-2} + a^{k-3}+···+a  + 1)$$
so we have
$$a^{k+1} - 1 = (a - 1)(a^k + a^{k-1} + a^{k-2} + a^{k-3}+···+a  + 1)$$
as desired.
 [/INST] To prove the induction step, you should prove that
$$a^{n+1}-1=(a-1)(a^n+a^{n-1}+\ldots+a+1),$$
given that
$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\ldots+a+1).$$
Of course this follows easily from the fact that
$$(a^{n+1}-1)-(a^n-1)=a^{n+1}-a^n=(a-1)a^n.$$
</s>