Sum of Powers of 2/Proof 2

Proof
Let $S \subseteq \N_{>0}$ denote the set of (strictly positive) natural numbers for which $(1)$ holds.

Basis for the Induction
We have:

So $1 \in S$.

This is our basis for the induction.

Induction Hypothesis
We now show that, if $k \in S$ is true, where $k \ge 1$, then it logically follows that $k + 1 \in S$.

So this is our induction hypothesis:
 * $\ds \sum_{j \mathop = 0}^{k - 1} 2^j = 2^k - 1$

Then we need to show:
 * $\ds \sum_{j \mathop = 0}^k 2^j = 2^{k + 1} - 1$

Induction Step
This is our induction step:

So $k \in S \implies k + 1 \in S$.

It follows by the Principle of Finite Induction that $S = \N_{>0}$.

That is:
 * $\ds \forall n \in \N_{> 0}: \sum_{j \mathop = 0}^{n - 1} 2^j = 2^n - 1$