Sum of k Choose m up to n

Theorem
Let $m, n \in \Z: m \ge 0, n \ge 0$.

Then:
 * $\displaystyle \sum_{k \mathop = 1}^n \binom k m = \binom {n + 1} {m + 1}$

where $\displaystyle \binom k m$ is a binomial coefficient.

Proof
Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition:
 * $\displaystyle \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$

Basis for the Induction
$\map P 0$ says:
 * $\dbinom 0 m = \dbinom 1 {m + 1}$

When $m = 0$ we have by definition:
 * $\dbinom 0 0 = 1 = \dbinom 1 1$

When $m > 0$ we also have by definition:
 * $\dbinom 0 m = 0 = \dbinom 1 {m + 1}$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P r$ is true, where $r \ge 0$, then it logically follows that $\map P {r + 1}$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{k \mathop = 1}^r \binom k m = \binom {r + 1} {m + 1}$

Then we need to show:
 * $\displaystyle \sum_{k \mathop = 1}^{r + 1} \binom k m = \binom {r + 2} {m + 1}$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall m, n \in \Z, m \ge 0, n \ge 0: \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$