Prove a combinatorics statement I’m reading a textbook about combinatorics and in the exercise section there is a question:
Prove that this statement is true .
$$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+...+\binom{n+m}{m}=\binom{n+m+1}{m}$$
I tried to expand every part with pascal theory which states:
$$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$$
Or use a couple of other theories but they took me nowhere 
I can really use a hint or an answer or even someone mentioning the theory I can use for solving this.
Thank you guys . you are amazing.
 A: You just have to replace $n\choose 0$ by $n+1\choose 0$ because value of both the terms is 1.
The reason behind this is that, the moment we write $n+1\choose 0$, it will get added up in the next term as $${n+1\choose 0}+{n+1\choose 1}={n+2\choose 1}$$
Now $n+2\choose 1$ will get added in the next term and so on. 
A: If I'm not mistaken induction on $m$ using $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$ should work. But there is probably a very nice combinatoric proof involving a bijection between two sets of exactly thoses sizes.
A: You can also perform induction on $m$. For $m=1$ the property reads
$$
\binom{n}{0} + \binom{n+1}{1}=\binom{n+2}{1},
$$
and this is obviously true (just compute each term). We just need to check that
$$
\sum_{k=0}^m \binom{n+k}{k} = \binom{n+m+1}{m} \Rightarrow \sum_{k=0}^{m+1} \binom{n+k}{k} = \binom{n+m+2}{m+1}.
$$
Now,
$$
\sum_{k=0}^{m+1} \binom{n+k}{k} = \sum_{k=0}^{m} \binom{n+k}{k} + \binom{n+m+1}{m+1}=  \binom{n+m+1}{m} + \binom{n+m+1}{m+1}=\binom{n+m+2}{m+1}.
$$
