Rising Sum of Binomial Coefficients/Proof by Induction

Proof
Proof by induction:

Let $n \in \Z$.

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

$\map P 0$ is true, as this just says:
 * $\dbinom n n = \dbinom {n + 1} {n + 1}$

But $\dbinom n n = \dbinom {n + 1} {n + 1} = 1$ from the.

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

So:
 * $\displaystyle \sum_{j \mathop = 0}^1 \binom {n + j} n = \binom {n + 2} {n + 1}$

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

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\displaystyle \sum_{j \mathop = 0}^k \binom {n + j} n = \binom {n + k + 1} {n + 1}$

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

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$

Finally, from Symmetry Rule for Binomial Coefficients:
 * $\dbinom {n + m + 1} {n + 1} = \dbinom {n + m + 1} m$

Proof

 * : $\text {3-1}$ Permutations and Combinations: Exercise $5$