Is collapsing considered a legitimate proof?

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:

$$\begin{align}S(0) &= S(k) \\&= 2^k + 2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1} \\&= 2\cdot2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1} \\&= 2^{k+1} + 2^{k+1} + 2^{k+2} + ... + 2^{n-1}\\&=S(k+1)\end{align}$$

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.

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.