Proof.

We must prove that \(1 + 2 + 2^2 + 2^3 + \cdots +2^n = 2^{n+1} - 1\) for all \(n \in \N\text{.}\) Thus let \(P(n)\) be the statement \(1 + 2 + 2^2 + \cdots + 2^n = 2^{n+1} - 1\text{.}\) We will prove that \(P(n)\) is true for all \(n \in \N\text{.}\) First we establish the base case, \(P(0)\text{,}\) which claims that \(1 = 2^{0+1} -1\text{.}\) Since \(2^1 - 1 = 2 - 1 = 1\text{,}\) we see that \(P(0)\) is true. Now for the inductive case. Assume that \(P(k)\) is true for an arbitrary \(k \in \N\text{.}\) That is, \(1 + 2 + 2^2 + \cdots + 2^k = 2^{k+1} - 1\text{.}\) We must show that \(P(k+1)\) is true (i.e., that \(1 + 2 + 2^2 + \cdots + 2^{k+1} = 2^{k+2} - 1\)). To do this, we start with the left-hand side of \(P(k+1)\) and work to the right-hand side:

\begin{align*} 1 + 2 + 2^2 + \cdots + 2^k + 2^{k+1} = \amp ~ 2^{k+1} - 1 + 2^{k+1} \amp \text{by inductive hypothesis}\\ = \amp ~2\cdot 2^{k+1} - 1 \amp\\ = \amp ~ 2^{k+2} - 1 \amp \end{align*}

Thus \(P(k+1)\) is true so by the principle of mathematical induction, \(P(n)\) is true for all \(n \in \N\text{.}\)