proving combinatorics identity - $\sum_{k=0}^m{n-k \choose m-k}={n+1 \choose m}$ 
Prove that for every $n \ge m \ge 1 , \sum_{k=0}^m{n-k \choose m-k}={n+1 \choose m}$ 

I've tried saying that the RHS represents the number of binary series with m "1" 's and n+1-m "0"'s, but I couldn't find out what k represents at the LHS.
Thanks

$$
\begin{align}
\sum_{k=0}^m\binom{n-k}{m-k}
&=\sum_{k=0}^m\binom{n-k}{n-m}\tag{1}\\
&=\binom{n+1}{n-m+1}\tag{2}\\[3pt]
&=\binom{n+1}{m}\tag{3}
\end{align}
$$
Explanation:
$(1)$: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: Equation $(9)$ from this answer
$(3)$: $\binom{n}{k}=\binom{n}{n-k}$
