Combo Identity: How to prove this using Induction 
$$ \sum_{n = 0}^{\infty} \binom{n + k}{k}x^n = \dfrac{1}{(1 - x)^{k + 1}} $$

Could someone suggest how I should get started to prove this using induction?
 A: HINT:
$$\frac1{(1-x)^{m+1}}=\frac{1-x}{(1-x)^{m+2}}=\frac1{(1-x)^{m+2}}-\frac x{(1-x)^{m+2}}$$
$$\implies(1-x)^{-(m+1)}=(1-x)^{-(m+2)}-x(1-x)^{-(m+2)}$$
Assume that the formula is true for $k=m+2$ and establish the same for $k=m+1$

Alternatively, use  $$(1-x)(1-x)^{-(m+2)}=(1-x)^{-(m+1)}$$
Assume that the formula is true for $k=m+1$ and establish the same for $k=m+2$
A: You start like that:
For $k=0$: $$\sum_{n=0}^{\infty} \binom{n}{n} x^n=\sum_{n=0}^{\infty} x^n=\frac{1}{1-x} \checkmark$$
Then,we suppose that: $$\sum_{n=0}^{\infty} \binom{n+k}{k} x^n=\frac{1}{(1-x)^{k+1}}$$
For $k+1$:
$$\sum_{n=0}^{\infty} \binom{n+k+1}{k+1} x^n=\sum_{n=0}^{\infty} \frac{n+k+1}{k+1} \binom{n+k}{k}x^n$$
then you have to continue,in order to find a relation that stands.
A: Why do you need proof by induction here? Use the Binomial Theorem! Simply set $-k-1 = v$:
$$
(1-x)^{v} = \sum_{j=0}^{\infty}\binom{v}{j}(-x)^j
$$
Now use the property of the binomial coefficient: $\binom{v}{j} = v (v-1) \ldots (v-j+1)\cdot \frac{1}{j!}$ Now use the definition for $v$: $\binom{-k-1}{j} = (-1)^j \frac{(k+1)(k+2) \ldots (k+j-1)}{j!} = (-1)^j =\frac{(k+j-1)!}{k! j!} = \frac{
(-1)^j}{k} \binom{k+j-1}{j}$. The $(-1)^j$ term cancels out, and the expression becomes what you wanted to prove.
