TITLE: Is there a generalization (surely there is) of this simple combinatorial identity?
QUESTION [6 upvotes]: I was just doing some algebra on a paper and obtained: $$\sum_{l=0}^{n-1} {{n+l} \choose l}={2n \choose {n+1}}$$
Are there some generalizations of this identity?
One possible generalization would be $$F_m(n)=\sum_{l=0}^{n-1} {{n+l} \choose l}^{m}$$ but it is not in my reach to obtain what would $F_m(n)$ equal to for every $m \in \mathbb N$
Other possible generalization could be of the form $$G_k(n)=\sum_{l=0}^{n-1} k^l \cdot{{n+l} \choose l}$$
I am interested in any generalization(s), not neccessarily of the forms I mentioned (I could mention some other forms but so could you so there is no need to do that).
You do not need to prove in an answer a generalization that you mention but it would be nice if you would point me to a direction where that generalization is mentioned and proven.
I asked a same question on MSE and received neither a comment nor an answer so I deleted that question there and decided to ask it here, although this is a low-level question for MO.

REPLY [9 votes]: What about the generalised Vandermonde identity:
$$
{ n_1+\dots +n_p \choose m }= \sum_{k_1+\cdots +k_p = m} {n_1\choose k_1} {n_2\choose k_2} \cdots {n_p\choose k_p}
$$
See also Hockey-stick identity and Identities involving binomial coefficients.

REPLY [6 votes]: Since $\binom{n+l}{l} = \binom{n+l}n$, we have
$$\sum_{l=0}^{n-1} \binom{n+l}{l} = \sum_{i=n}^{2n-1} \binom{i}{n}.$$
Here the lower and upper summation bounds as well as the lower index of binomial coefficients depend on $n$. This is a particular case of a more general formula, where all three entities are independent:
$$\sum_{i=n}^m \binom{i}{k} = \binom{m+1}{k+1} - \binom{n}{k+1}.$$

REPLY [5 votes]: Formula 4.2.5.37 on page 503 in Prudnikov, Brychkov, Marichev is 
\begin{equation}
 \sum_{k=0}^n\binom{a+k}a\binom{b-k}{b-n}=\binom{a+b+1}n. 
\end{equation}
Substituting here $a=n+1$ and $b=n$, and then replacing $n$ by $n-1$, we get your identity. 
The book Prudnikov, Brychkov, Marichev is in Russian, but this should be no problem here.