Proving divisibility using induction Question:
Let $a$ be an integer different than $1$. Prove by induction, that for any $n \in\Bbb N$, $ a^n -1$ is divisible by $a-1$. 
My attempt:
Base Case is trivial.
I.H: Assume $ a^k -1$ is divisible by $a-1$, $k \in N$
Then,
$$ a^{k+1} -1 = a\cdot a^k -1 + a^k -1 - a^k + 1 = a\cdot a^k - a^k + a^k -1 = (a^k)(a-1) + (a^k -1)$$which is divisible by $ a-1$. 
Hence, proven by induction.
Is this the correct way to prove it using induction? Is there a more efficient way to prove it using induction?  
 A: Hint $\ $ The inductive step follows by telescopy. $ $ Let $f(k)= a^k-1$
Then $\,\color{#0a0}{f(k\!+\!1)\!-\!f(k)} = a^{k+1}-a^k = (a\!-\!1)a^k$ is divisible by $\,a\!-\!1$
so if $\,\color{#c00}{f(k)}$ is divisible by $\,a\!-\!1\,$ then so too is $\,f(k\!+\!1) = \color{#0a0}{f(k+1)\!-\!f(k)} + \color{#c00}{f(k)}$
Remark $\, $ Summing the above makes the telescopic cancellation explicit 
$$\begin{align}a^{k+1}-1 =\, &\,\ \ \color{#c00}{a^{k+1}\!-a^k}+\color{#0a0}{a^k\!-a^{k-1}} + \cdots+ a^1\!-a^0\\[.3em]
=\, &\,(a-1)\, (\color{#c00}{a^k} + \color{#0a0}{a^{k-1}}+\cdots+1),\ \ \text{or, more formally}\\ f(k\!+\!1)-f(0) =\,& \sum_{i=0}^k (f(i\!+\!1)\!-\!f(i)) = \sum_{i=0}^ka^i(a\!-\!1) = (a\!-\!1)\sum_{i=0}^k a^i\end{align} $$ 
The first equality above, expressing $f(k\!+\!1)-f(0)$ as the sum of its first differences, has a trivial (and obvious) proof by induction. Once you prove this general form inductively, you can use it as a lemma to inductively prove many special cases - which are ubiquitous. You can find many examples and much further discussion of telescopic induction in various prior posts.
A: Let $a>1$ be an integer. Prove by induction that for any $n \in \mathbb{N}$, $ a^n -1$ is divisible by $a-1$. 
$\textbf{Proof}.$ Base case: let $n=1$. Then it is clear that $a-1|a-1$. 
Induction hypothesis: assume that $a-1|a^k-1$ for some integer $k$. Consider $a^{k+1}-1$, which could be rewritten as 
$$
\begin{align*}
a^{k+1}-1 &= a\cdot a^k + a^k-a^k-1 \\ 
&= a^k(a-1)+ (a^k-1).  \\ 
\end{align*}
$$
Since $a-1|a^k(a-1)$ and $a-1|a^k-1$ by the induction hypothesis, $a-1|a^{k+1}-1$. 
Thus $a-1|a^n-1$.  
