Proving that $\sum\limits_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$ 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
 A: With minor modification, your opening up procedure will work. Write the Pascal Identity in the non-traditional order $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$. (We have just reversed the addition order of your two terms.)
Now we do your computation, for clarity with  $m+n+1=17$ and $m+1=5$. We get
$$\binom{17}{5}=\binom{16}{4}+\binom{16}{5}=\binom{16}{4}+\binom{15}{4}+\binom{15}{5}.$$
Now decompose $\dbinom{15}{5}$ as $\dbinom{14}{4}+\dbinom{14}{5}$, and then  $\dbinom{14}{5}$ as $\dbinom{13}{4}+\dbinom{13}{5}$, and so on.
If you want to be formal, you can easily write it up as an induction on $n$. 
A: $${m+i+1\choose m+1}={m+i\choose m+1}+{m+i\choose m} $$
$$\Rightarrow $$ $${m+i\choose m}={m+i+1\choose m+1}-{m+i\choose m+1}$$
$$\text{let} \ i=1,2,...n$$
$${m+1\choose m}={m+1+1\choose m+1}-{m+1\choose m+1}$$
$${m+i\choose m}={m+i+1\choose m+1}-{m+i\choose m+1}$$
$${m+n\choose m}={m+n+1\choose m+1}-{m+n\choose m+1}$$
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 
$$\sum_0^n{m+i\choose m}={m+n+1\choose m+1}$$
A: Here is a purely combinatorial proof:
Consider picking $m+1$ numbers out of $\{1,2,...,m, \color{  #009900}{m + 1}, \color{  #009900}{m + 1} + 1,...,\color{  #009900}{m + 1} + (n - 1),\color{  #009900}{m + 1}+n\}$.
The right hand side of your equation is clearly equal to the number of ways of doing this.
Now for any given choice of $m+1$ numbers, the highest number chosen must be some $k$ with $\color{  #009900}{m + 1} \leq k \leq \color{  #009900}{m + 1}+n$. In each of these cases, we must select the remaining $m$ numbers to be chosen from the $k-1$ numbers smaller than $k$.
For $k = m +1$, must pick $m$ numbers to the left of $m + 1$, out of $\{\color{  #0073CF}{1, 2, ..., m}, m+1\}$.
Since there are $ \color{#0073CF}{m}$ such numbers, so $\color{#0073CF}{m}$ possible choices for $m$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{m}}{m}$. 
For $k = m +2$, must pick $m$ numbers to the left of $m + 2$, out of $\{\color{  #0073CF}{1, 2, ..., m, m +1}, m+2\}$.
Since there are $ \color{#0073CF}{m + 1}$ such numbers, so $\color{#0073CF}{m + 1}$ possible choices for $m$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{m + 1}}{m}$. 
...
For $k = m + 1 + n$, must pick $m$ numbers to the left of $m + 1 + n$, out of $\{\color{  #0073CF}{1, 2, ..., m, m +1, ..., m + n}, m+ n + 1\}$.
Since there are $ \color{#0073CF}{m + n}$ such numbers, so $\color{#0073CF}{m + n}$ possible choices for $m$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{m + n}}{m}$. 
Summing up the number of ways of doing this for $k=m+1,...,m+n+1$ yields the LHS of your equation.
