Find a combinatorial proof for $\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + ... + \binom{n-k}{0}$ 
Let $n$ and $k$ be integers with $n \geq k \geq 0$. Find a combinatorial proof for
$$\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + \cdots + \binom{n-k}{0} .$$

My approach:
I was thinking to use the binomial formula as in $$2^n = \sum{\binom{n}{k}1^k1^{n-k}} .$$
I also tried to use Pascal's Identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$.

It follows from Pascal's identity that\begin{align}\binom{n+1}k&=\binom nk+\binom n{k-1}\\&=\binom nk+\binom{n-1}{k-1}+\binom{n-1}{k-2}\\&=\binom nk+\binom{n-1}{k-1}+\binom{n-2}{k-2}+\binom{n-2}{k-3}\\&=\cdots\end{align}Can you take it from here?
