We proceed by induction on $N$. The base case of $N = 0$ is immediate since
\begin{equation}
\sum_{n = 0}^0 2^n
= 2^0
= 1
= 2^1 - 1
= 2^{0 + 1} - 1
\end{equation}
As an induction hypothesis, suppose then that the claim holds for some positive integer $N \geq 0$. We have
\begin{equation}
\begin{split}
\sum_{n = 0}^{N + 1} 2^n
= 2^{N + 1} + \sum_{n = 0}^N 2^n
= 2^{N + 1} + 2^{N + 1} - 1
= 2 \cdot 2^{N + 1} - 1
= 2^{N + 2} - 1
\end{split}
\end{equation}
This shows that the claim holds for $N + 1$. The result is now a consequence of
R800: Proof by principle of weak mathematical induction. $\square$
