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 $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$.

A combinatorial proof based upon lattice paths.

The number of lattice paths of length $n+1$ from $(0,0)$ to $Q_k=(n-k+1,k)$ consisting of $(1,0)$-steps and $(0,1)$-steps is $$\binom{n+1}{n-k+1}=\binom{n+1}{k}$$ since we have to choose precisely $n-k+1$ $(1,0)$-steps from a total of $n+1$ steps.
                                
We observe that each path which goes from $(0,0)$ to $Q_k$ crosses the vertical line $x=n-k$. We can therefore partition all $\binom{n+1}{k}$ paths according to points $P_j=(n-k,j), 0\leq j\leq k$ which are followed by a horizontal step to $(n-k+1,j)$ and the vertical steps which lead to $Q_k$.
Since there are precisely $\binom{n-k+j}{j}$ paths from $(0,0)$ to $P_j$ the number of all paths from $(0,0)$ to $Q_r=(n-k+1,k)$ is
  \begin{align*}
\color{blue}{\binom{n+1}{k}=\binom{n}{k}+\binom{n-1}{k-1}+\cdots+\binom{n-k}{0}}
\end{align*}

