TITLE: Proving that $\sum\limits_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$
QUESTION [9 upvotes]: I have to prove that:

$$\sum_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$$

I tried to open up the right side with Pascal's definition that:
$$ { n \choose k} = {n-1 \choose {k}} + {n-1 \choose {k-1}}$$
Here is what I came up with, and I am sure it is wrong because it does not equal the left side:
$$ {m+n+1 \choose m+1} = {m+n \choose m+1} + {m+n \choose m} = ... ={m+n \choose m+1} + {m+n-1 \choose m} + {m+n-2 \choose m-1} + ... + {m \choose m+1} = \sum_{k=0}^{n} {m+k \choose m+k-n+1 }  $$
Which, again, probably is wrong because it is not equal $\sum_{k=0}^{n} { m+k \choose m}$.
Any help is appreciated

REPLY [2 votes]: $${m+i+1\choose m+1}={m+i\choose m+1}+{m+i\choose m} $$
$$\Rightarrow $$ $${m+i\choose m}={m+i+1\choose m+1}-{m+i\choose m+1}$$
$$\text{let} \ i=1,2,...n$$
$${m+1\choose m}={m+1+1\choose m+1}-{m+1\choose m+1}$$
$${m+i\choose m}={m+i+1\choose m+1}-{m+i\choose m+1}$$
$${m+n\choose m}={m+n+1\choose m+1}-{m+n\choose m+1}$$
Add all the terms and we have following identities
$$\sum_1^n{m+i\choose m}={m+n+1\choose m+1}-{m+1\choose m+1}$$
And we know 
$${m+1\choose m+1}={m\choose m}$$
So 
$$\sum_0^n{m+i\choose m}={m+n+1\choose m+1}$$