Simple binomial theorem proof: $\sum_{j=0}^{k} \binom{a+j}j = \binom{a+k+1}k$ I am trying to prove this binomial statement:

For $a \in \mathbb{C}$ and $k \in \mathbb{N_0}$, $\sum_{j=0}^{k} {a+j \choose j} = {a+k+1 \choose k}.$ 

I am stuck where and how to start. My steps are these: 
${a+j \choose j} = \frac{(a+j)!}{j!(a+j-j)!} = \frac{(a+j)!}{j!a!}$ but now I dont know how to go further to show the equality.
Or I said:   $\sum_{j=0}^{k} {a+j \choose j} = {a+k \choose k} +\sum_{j=0}^{k-1} {a+j \choose j} = [help!] = {a+k+1 \choose k}$
Thanks for help!

Look for Pascal's rule "http://en.wikipedia.org/wiki/Pascal's_rule". After this it is easy.
