Is collapsing considered a legitimate proof? For example if I want to prove that $2^n - 1 = 1 + 2 + 4 + 8 +...+ 2^{n-1}$ I can obviously use induction and that is accepted. But I can also collapse it like:
To Prove $2^n = S(n)$:


*

*$S(n) = (1 + 1) + 2 +...+ 2^{n-1}$

*$S(n) = (2 + 2) + 4 + 8 +...+2^{n-1}$

*$S(n) = (4 + 4) + 8 +...+2^{n-1}$


and so on until
$S(n) = 2^{n-1} + 2^{n-1} = 2^n$
Is this method of collapsing considered a legitimate and presentable proof? 
 A: This depends very much on the degree of rigour you want.
If you want to be really rigorous, or need to, then you would need to formalize “so on” by showing by induction that
$$
 1 + \sum_{i=0}^{n-1} 2^i
 = 2^k + \sum_{i=k}^{n-1} 2^i
 \quad
 \text{for all $k = 0, \dotsc, n$},
$$
which shows the statement for $k = n$.
If you are satisfied with less rigour, and I think most (read: nearly all) mathematicians will be in a case like this, then “so on” is good enough, as the idea behind your proof is pretty clear. It is, however, important that you are able to give a formal proof.
A: Well, sort of, but in fact, writing proofs like the one you want to write is why induction exists. Whenever people say something like "and so on until", they're expressing your intuition that it's possible to continue the argument by induction. The whole point of the method of induction is to make intuitions like this one precise.
Let $S(k)=2^k + 2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1}$. Then what we want to show is that $S(0) = 2^n$. Your proof basically amounts to saying $S(0) = S(1) = S(2) = ...$ "and so on", until we get $S(0) = S(n-1)$. Notice that $S(n-1) = 2^{n-1} + 2^{n-1}$, which obviously equals $2^n$. So we get $S(0) = 2^n$. To phrase this as a proof by induction, we're going to prove by induction that $S(0) = S(k)$ for all $k<n$, thus we'll obtain $S(0)=S(n-1)$ at the end.
Obviously, $S(0) = S(0)$. Now suppose $S(0) = S(k)$. Then:
$$\begin{align}S(0) &= S(k) \\&= 2^k + 2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1} \\&= 2\cdot2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1} \\&= 2^{k+1} + 2^{k+1} + 2^{k+2} + ... + 2^{n-1}\\&=S(k+1)\end{align}$$
A: Of course: for every finite $n$ you have absolutely convergent series which allows you to manipulate it's memebers in a way you are doing it. Although it seems if you want to present you proof in a most formal way, induction will take up less space.
A: the consensus here seems to be "yeah, the proof is ok but rather informal since it invokes induction 'coded' in the 'and-so-on' statement". I don't agree. Imho this is no proof at all and there is no induction argument given.


*

*The Argument starts with the claim (short of subtracting one from both sides of the equation). Why bother and continue?

*The modifications made in order to get to the next line are completely unclear to me. Are all the elements in the series doubled? Then the result (left side) should be doubled as well but it isn't?

