Another combinatorial proof

Give combinatorial or bijective proofs of the following. Part of your job is to determine all values of n,k, and/or m for which the identities are valid.

c) $\displaystyle \left({0\atop m}\right)+\left({1\atop m}\right)+\left({2\atop m}\right)+\ldots+\left({n\atop m}\right)=\left({n+1\atop m+1}\right)$

I can't get very far with this one either. I notice that all terms on the left hand side are zero until we get to $\displaystyle \left({m\atop m}\right)$. Then I calculate all terms equal 1, so we have n-m terms which makes the left hand side equal n-m.

The right hand side I get $\displaystyle \frac{(n+1)!}{(m+1)!(n+1-m-1)!}=\frac{(n+1)!}{n!(m+1)}=\frac{n+1}{m+1}$

I can get no further.

Can someone help or tell me if I have the whole combinatorial proof thing wrong.