combinatorial proof for an identity with generating functions $\frac{1}{(1-x)^n} = \sum_{r=0}^\infty{r+n-1 \choose r}x^r$

Question

i'm trying to prove $\frac{1}{(1-x)^n} = \sum_{r=0}^\infty{r+n-1 \choose r}x^r$ with a combinatorial proof using integer solution problems and generating functions, but I can't think of any integer solution problem that would work as the combinatorial situation for this identity. Any help would be appreciated!

Answer 1

$$\frac{1}{1-x}=\sum_{p=0}^{+\infty}x^p$$ Thus using Cauchy product : $$ \frac{1}{(1-x)^n}=\sum_{p=0}^{+\infty}a_{n,p}x^p $$ with $$a_{n,p}=\sum_{\underset{k_1+\ldots+k_n=p}{(k_1,\ldots,k_n)\in\mathbb{N}^n}}1=\text{Card}\{(k_1,\ldots,k_n)\in\mathbb{N}^n\ |\ k_1+\ldots+k_n=p\}$$ Moreover $$a_{n,p}=\sum_{k=0}^p a_{n-1,k}$$ (make a partition on the value of $k_n\in[\![0,p]\!]$). We show by induction on $n$ that $a_{n,p}=\binom{n+p-1}{p}$ : the case $n=1$ is trivial because $a_{1,p}=1$ for all $p$. Let $n\in\mathbb{N}^*$ such that $a_{n,p}=\binom{n+p-1}{p}$ for all $p$, we have $$ a_{n+1,p}=\sum_{k=0}^p a_{n,k}=\sum_{k=0}^p\binom{n+k-1}{k}=\sum_{k=0}^p \binom{n+k}{k}-\binom{n+k-1}{k-1}=\binom{n+p}{p} $$