Let $N \in \mathbb{N}$ be a D996: Natural number.

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. $\square$