Summation Formula for Reciprocal of Binomial Coefficient/Proof 1

Proof
First note that:

The proof continues by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom n k \dfrac {\left({-1}\right)^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$

Basis for the Induction
$P \left({0}\right)$ is the case:

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({m}\right)$ is true, for all $m \ge 0$, then it logically follows that $P \left({m + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom m k \dfrac {\left({-1}\right)^k} {k + x} = \dfrac 1 {x \binom {m + x} m}$

from which it is to be shown that:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom {m + 1} k \dfrac {\left({-1}\right)^k} {k + x} = \dfrac 1 {x \binom {m + 1 + x} {m + 1} }$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \Z_{\ge 0}: \displaystyle \sum_{k \mathop \ge 0} \binom n k \dfrac {\left({-1}\right)^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$