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