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 "off by one errors? E.g. $n = ab$ vs $n = ab -1$?
To prove:
For $a, b > 1$
$2^{ab} - 1$ 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 $2^{ab} - 1$.
$\square$

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 $x=2^a$
