Sum of k Choose m up to n

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

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

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

Proof
Proof by induction:

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

Basis for the Induction
$P(0)$ says $\displaystyle \binom 0 m = \binom 1 {m + 1}$.

When $m = 0$ we have by definition $\displaystyle \binom 0 0 = 1 = \binom 1 1$.

When $m > 0$ we also have by definition $\displaystyle \binom 0 m = 0 = \binom 1 {m+1}$.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r+1}\right)$ is true.

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

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

Induction Step
This is our induction step:

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

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