Proof that $2^{ab} - 1$ is divisible by $2^a - 1$ Could you please comment on the correctness and quality of the following proof?
I'm a bit suspicious of the line $n = ab, a = 1, r = 2^a$. Should it be $r = 2$? And are there any &quot;off by one errors? E.g. $n = ab$ vs $n = ab -1$?
To prove:
For $a, b &gt; 1$
$2^{ab} - 1$ is divisible by $2^a - 1$
Explanation:
$2^{ab} - 1$ can be expressed as $\sum_{i=0}^{ab-1}2^i$
Considering the formula for the sum of a geometric series
$S_n = a \frac{r^n-1}{r-1}$
with
$n = ab, a = 1, r = 2^{a}$
we have
$S_n = \frac{(2^a)^b-1}{2^a-1}$
From which it can be seen that $2^a - 1$ is indeed a factor of $2^{ab} - 1$.
$\square$
I'm going to be picky:
You state that $2^{ab}-1=\sum_{k=0}^{ab-1} 2^k$ but never actually use it. So leave it out; it just confuses things.
You should state explicitly that $S_n = \sum_{k=0}^{b-1} (2^a)^k$; as it is written I use confusing it with $\sum_{k=0}^{ab-1}=2^{ab}-1$ which is and entirely different completely irrelevant sum-- I was confused by the first line.
Then I'd like the, admittedly trivial but key, statement that: $S_n$ is an integer (as a finite sum of integers).

I.E. the proof should be:  $S_n= \sum_{k=0}^{b-1} (2^a)^k$ is integer. But $S_n = \frac {(2^a)^b -1}{2^a - 1}$.  So $2^a - 1| (2^a)^b - 1= 2^{ab}-1$.

Otherwise the proof is good.
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Although......
For the purpose of a stronger proof.
If we can prove that $x-1$ always divides $x^n - 1$ we will be done by letting $x = 2^a$ and $n = b$.
And the formula for the geometric series states precisely that:  That $\sum_{k=0}^{n-1} x^k$ which is clearly an integer is equal to $ \frac{x^n-1}{x-1}$.  So $x-1|x^n -1$.
Of course, it could be that this exercise is expecting you to prove the formula for a geometric series.
Which you should be prepared to do.
Pf:  $(x- 1)(\sum_{k=0}^{n-1} x^k) = x\sum_{k=0}^{n-1}x^k - \sum_{k=0}^{n-1} x^k=$
$\sum_{k=0}^{n-1} x^{k+1} - \sum_{k=0}^{n-1} x^k=$
$\sum_{k=1}^n x^k - \sum_{k=0}^{n-1} x^k=$
$ ((\sum_{k=1}^{n-1} x^k) + x^n)- (x^0 + \sum_{k=1}^{n-1} x^k)=$
$x^n - 1$.
So $\sum_{k=0}^{n-1} x^k = \frac {x^n - 1}{x-1}$.
(Assuming, of course, $x \ne 1$... Obviously if $x = 1$ then $(1-1)\sum_{k=0}^{n-1} 1^k= 0*n = 0 =1^n-1$ which is fairly pointless)
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More interesting is proving that if $2^m -1$ is not divisible by $2^a-1$ unless $m$ is divisible by $a$.
If we pick the common ratio to be $2^a$,
\begin{align}
\sum_{i=0}^{ab-1}2^i = \sum_{t=0}^{b-1}\sum_{i=0}^{a-1}2^{at+i} =\sum_{t=0}^{b-1}(2^a)^t \left(\sum_{i=0}^{a-1}2^i\right)
\end{align}
the first term is $\left(\sum_{i=0}^{a-1}2^i\right)$ and we have $b$ terms in total.
That is $$2^{ab}-1=\sum_{t=0}^{b-1}(2^a)^t \left(2^a-1\right)$$
$$\frac{2^{ab}-1}{2^a-1}=\sum_{t=0}^{b-1}(2^a)^t$$
Hence $2^a-1 | 2^{ab}-1$
I usually prove it in the following way:

$x^b-1$ is clearly divisible by $x-1$ since $1^b-1=0$.

Now, we just have to let $x=2^a$