Answer
See below.
Work Step by Step
Proofs using mathematical induction consists 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 we prove that then the statement also holds for $n + 1$.
Hence here: 1) For $n=1: 2=2^{1+1}-2$.
2) Assume for $n=k: 2+4+...+2^{k}=2^{k+1}-1$. Then for $n=k+1$: $1+2+...+2^{k}+2^{k+1}=2^{k+1}-1+2^{k+1}=2\cdot2^{k+1}-1=2^{k+2}-1$
Thus we proved what we wanted to.