Answer
See below.
Work Step by Step
Proofs using mathematical induction consist of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then prove that the statement also holds for $n + 1$.
Hence here:
1) For $n=1: 2^0=2^1-1$.
2) Assume for $n=k: 1+2+...+2^{k-1}=2^k-1$. Then for $n=k+1$:
$1+2+...+2^{k-1}+2^k=2^k-1+2^k=2(2^k)-1=2^{k+1}-1.$
Thus we proved what we wanted to.
