What is the most elegant way to obtain $$\sum_{\beta \leq \alpha} \binom{\alpha}{\beta} = 2^{|\alpha|}$$ from the multinomial theorem, i.e. $$ (x_1+x_2+\ldots+x_n)^N = \sum_{|\alpha|=N}\frac{N!}{\alpha!}x^\alpha? $$ I was thinking of taking $n=2$ and $x_1=x_2 = 1$ to obtain $$ 2^N = \sum_{j+k = N} \frac{N!}{j!k!}, $$ and then setting $N=|\alpha|$ so that $$ 2^{|\alpha|} = \sum_{j+k=|\alpha|} \frac{|\alpha|!}{j!k!} $$ but I don't see how I can get the multi-index factorials from here. Any hints/better ways to do this?
EDIT: $\alpha$ and $\beta$ are multi-indices, see Multi-index notation
$\binom{\alpha}{\beta} = \prod_{i\leq n} \binom{\alpha_i}{\beta_i}$ so $\sum_{\beta\leq \alpha} \binom{\alpha}{\beta} =\sum_{\beta\leq \alpha} \prod_{i\leq n} \binom{\alpha_i}{\beta_i}$
Just notice how you can handle each $i$ separately to use the standard formula for integers: $\sum_{k\leq n} \binom{n}{k} = 2^n$.
Thus: $$ \sum_{\beta\leq \alpha} \binom{\alpha}{\beta} = \prod_{i\leq n} \sum_{\beta_i\leq \alpha_i}\binom{\alpha_i}{\beta_i}=\prod_{i\leq n}2^{\alpha_i}=2^{|\alpha|} $$