Sum of products of combinatorials In the proof of some proposition, it appears that the following statement should hold:

$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} =2^{2n}. $$

However, using the definition of combinatorials does not help:
$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} =  \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} \frac{(n+1)!}{(n+1-r)!r!} \frac{n!}{(n-\beta)! \beta!}.$$
I suspect that this has something to do with the binomial expansion. Any ideas?

We have
$$\sum_{r=1}^{n+1} {n+1\choose r}
\sum_{\beta=0}^{r-1} {n\choose \beta}
\\ = \sum_{r=1}^{n+1} {n+1\choose r}
\sum_{\beta\ge 0} {n\choose \beta} [[0\le \beta\le r-1]]
\\ = \sum_{r=1}^{n+1} {n+1\choose r}
\sum_{\beta\ge 0} {n\choose \beta}
[z^{r-1}] \frac{z^{\beta}}{1-z}
\\ = \sum_{r=1}^{n+1} {n+1\choose r}
[z^{r-1}] \frac{1}{1-z}
\sum_{\beta\ge 0} {n\choose \beta} z^{\beta}
\\ = \mathrm{Res}_{z=0} \frac{1}{1-z} (1+z)^n
\sum_{r=1}^{n+1} {n+1\choose r}
\frac{1}{z^r}
\\ = \mathrm{Res}_{z=0} \frac{1}{1-z} (1+z)^n
\left(-1 + \left(1+\frac{1}{z}\right)^{n+1}\right)
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1-z} (1+z)^{2n+1}
\\ = \sum_{q=0}^n {2n+1\choose q} = \frac{1}{2} 2^{2n+1}
= 2^{2n}.$$
