How to prove that $2^{n-1}=1+1+2+4+\dotsb+2^{n-2}$ for $n>1$? This is a claim in a textbook that I'm using, although it was not proven.
When trying this for small $n$'s it is clear.
\begin{align}
2^2&=1+1+2^1=4\\
2^3&=1+1+2+2^2\\
2^4&=1+1+2+4+2^3
\end{align}
What I was thinking for the proof was $2^{n-1}=2(2^{n-2})=2^{n-2}+2^{n-2}$, but obviously that's not incorporating addition, so I don't know where to go on. Is it just expressing $2^{n-2}$ as the first summation?

Let $S:=1+2+4+\dots+2^{n-2}$.
Then, since it's a geometric series, we have $S=\dfrac{1-2^{n-1}}{1-2}=2^{n-1}-1\implies S+1=2^{n-1}$.
