Sum of Geometric Sequence/Proof 4/Lemma

Lemma
Let $n \in \N_{>0}$.

Then:
 * $\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$

Proof
Proof by induction on $n$:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$

Basis for the Induction
$\map P 1$ is the case:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

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

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

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$