Combinatoric Proof of $\sum_0^n\binom{k-1+i}{k-1} = \binom{n+k}{k}$ $\sum_0^n\binom{k-1+i}{k-1} = \binom{n+k}{k}$
I think there are two different ways to prove the above identity: one is algebraic and the other one is combinatoric.
I have seen there's some ways to handle this problem with some sort of lattice like structure.
Any advice for approaching this problem in "combinatoric" way?
 A: Suppose $x_1,x_2,...\in \mathbb{N}\cup \{0\}$
and we need to answer numbers of solution of this nequality 
$$x_1+x_2+x_3+...+x_k\leq n$$ one method is partitioning ,so we have 
$$x_1+x_2+x_3+...+x_k\leq n  (\equiv)
\\\to\begin{cases}x_1+x_2+x_3+...+x_k = 0 & \left(\begin{array}{c}0+k-1\\ k-1\end{array}\right)\\x_1+x_2+x_3+...+x_k = 1 & \left(\begin{array}{c}1+k-1\\ k-1\end{array}\right)\\x_1+x_2+x_3+...+x_k =2&\left(\begin{array}{c}2+k-1\\ k-1\end{array}\right)\\\vdots\\x_1+x_2+x_3+...+x_k = n& \left(\begin{array}{c}n+k-1\\ k-1\end{array}\right)\end{cases} $$sum of the is $$\sum_0^n\binom{k-1+i}{k-1} $$ second method to find the number of solution is to add $\bf{x_{k+1}}$ as new variable ,and convert inequality to equation .
  so 
$$x_1+x_2+x_3+...+x_k \leq n  \space (\equiv)\\
x_1+x_2+x_3+...+x_k +\color{red} {\bf{x_{k+1}}}=n \to \binom{n+(k+1-1}{(k+1)-1}$$ hence
$$\sum_0^n\binom{k-1+i}{k-1} =\binom{n+(k+1)-1}{(k+1)-1}=\binom{n+k}{k}$$
A: Combinatorial Approach
The Stars and Bars Formula says that there are $\binom{k-1+i}{k-1}$ ways to put $i$ stars into $k$ bins. Thus, there are
$$
\sum_{i=0}^n\binom{k-1+i}{k-1}
$$
$$
\underbrace{\star\star\star\,\mid\,\star\,\mid\,\star\star}_\text{$i$ stars and $k-1$ bars}\qquad\underbrace{\star\star\star\star\star}_\text{$n-i$ stars}
$$
ways to put up to $n$ stars into $k$ bins. We can also count this same number by adding a bin to hold the excess $n-i$ stars and get that the number of ways to put the $n$ stars into the $k+1$ bins to be
$$
\binom{n+k}{k}
$$
$$
\underbrace{\star\star\star\,\mid\,\star\,\mid\,\star\star\,\mid\,\star\star\star\star\star}_\text{$n$ stars and $k$ bars}
$$

Algebraic Approach
$$
\begin{align}
\sum_{i=0}^n\binom{k-1+i}{k-1}
&=\sum_{i=0}^n\binom{k-1+i}{i}\binom{n-i}{n-i}\tag{1}\\
&=(-1)^n\sum_{i=0}^n\binom{-k}{i}\binom{-1}{n-i}\tag{2}\\
&=(-1)^n\binom{-k-1}{n}\tag{3}\\
&=\binom{k+n}{n}\tag{4}\\
&=\binom{k+n}{k}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $\binom{n}{k}=\binom{n}{n-k}$ and $\binom{k}{k}=[k\ge0]$
$(2)$: convert to negative binomial coefficients
$(3)$: Vandermonde's Identity
$(4)$: convert from negative binomial coefficients
$(5)$: $\binom{n}{k}=\binom{n}{n-k}$
