Combinatorial proof using list counting

Question

I'm having a lot of trouble coming up with the situation/question for a combinatorial proof that asks to prove $$2^0 + 2^1 + \dotsb + 2^{n−1} = 2^n − 1$$

Using list counting. The question I came up with for the right hand side is: what is the number of lists length $n$ you can form with $2$ inputs, excluding the list where all the inputs are $2$.

Answer 1

Hint: Think about binary strings of length $n$ with at least one $1$ ... & where the first $1$ occurs in the string.

Answer 2

Every list of $n$ $0$'s and $1$'s has $i$ leading $0$'s where $0 \leq i \leq n$. There is $1$ list with $n$ leading $0$'s. And in general there are $2^k$ lists with $n-k$ leading $0$'s. The total number of lists of $n$ $0$'s and $1$'s is $2^n$.