Alternating Sum and Difference of r Choose k up to n/Proof 2

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\displaystyle \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$

$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

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

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_{k \mathop \le m} \paren {-1}^k \binom r k = \paren {-1}^m \binom {r - 1} m$

from which it is to be shown that:
 * $\displaystyle \sum_{k \mathop \le m + 1} \paren {-1}^k \binom r k = \paren {-1}^{m + 1} \binom {r - 1} {m + 1}$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \Z_{\ge 0}: \displaystyle \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$