Consider paths from $(0,0)$ to $(k,n)$ that increase one of the two components on each step, so they make $n+k$ steps, among which $k$ horizontal (increasing $i$ in $(i,j)$) and $n$ vertical (increasing $j$).
There are $\binom{n+k}k$ such paths.
Let the first point with first coordinate $i$ be $(i,j_i)$ for $i=1,\ldots,k$, then clearly $0\leq j_1\leq\cdots\leq j_k\leq n$. Moreover all such sequences of values are allowed, and they determine the path: the horizontal steps are precisely those from $(i-1,j_i)$ to $(i,j_i)$ for $i=1,\ldots,k$.
Other points of view. If you take $y_i$ to be the index of the horizontal step to $(i,j_i)$ among all steps (which are numbered from $1$ to $n+k$), you'll get your $0. If you put a candy (star) at each point $(i,j)$ of the path and think of each horizontal step as a separation (bar), you'll have split your $n+k+1$ candies into $k+1$ vertical groups (for $i=0,1,\ldots,k$) the last group $i=k$ being the left-overs. But actually you want each group to have as many candies as it has vertical steps, not points, so you remove one from each group to correct this fencepost error, leaving groups of vertical steps of sizes $d_0,d_1,\ldots,d_k$ (each $d_i\geq0$), with $d_0+\cdots+d_k=n$. In the end you might just as well take $n+k$ steps, choose $k$ of them to be horizontal (bars) and the remaining $n$ to be vertical (stars). But I think I already said something similar at the beginning.