What makes this equation equal to the combination equation I saw this statement: $${n \choose r}=\sum_{n_{r-2}=0}^{\left(n+1-r\right)}\left(\sum_{n_{r-3}=0}^{\left(n+1-r\right)-n_{r-2}}\left(\sum_{n_{r-4}=0}^{\left(n+1-r\right)-\left(n_{r-3}+n_{r-2}\right)}\left(...\left(\sum_{n_{1}=0}^{\left(n+1-r\right)-\sum_{k=2}^{r-2}n_{k}}\left(\sum_{n_{0}=0}^{\left(n+1-r\right)-\sum_{k=1}^{r-2}n_{k}}\left(n_{0}\right)\right)\right)\right)\right)\right)$$ somewhere, and I was curious, why is this statement true?
 A: The index  $n_{r-2}$ of the   left-most   sum indicates  to consider showing the  validity of the  sum   for  integral $n\geq  r\geq    2$. In  order to see what's going on we look at first at a  small  example, let's say $r=4$. This seems to be small enough to be manageable, but should be large enough to see the essential steps.

Case $r=4$:
We obtain
\begin{align*}
\color{blue}{\sum_{n_2=0}^{n-3}\sum_{n_1=0}^{n-3-n_2}\sum_{n_0=0}^{n-3-n_2-n_1}n_0}
&=\sum_{n_2=0}^{n-3}\sum_{n_1=0}^{n_2}\sum_{n_0=0}^{n_2-n_1}n_0\tag{1}\\
&=\sum_{n_2=0}^{n-3}\sum_{n_1=0}^{n_2}\sum_{n_0=0}^{n_1}n_0\tag{2}\\
&=\sum_{n_2=1}^{n-3}\sum_{n_1=1}^{n_2}\sum_{n_0=1}^{n_1}n_0\tag{3}\\
&=\sum_{n_2=1}^{n-3}\sum_{n_1=1}^{n_2}\sum_{n_0=1}^{n_1}\sum_{n_{-1}=1}^{n_0}1\tag{4}\\
&=\sum_{1\leq  n_{-1}\leq n_0\leq n_1\leq n_2\leq n-3}1\tag{5}\\
&=\binom{4+(n-3)-1}{4}\tag{6}\\
&\,\,\color{blue}{=\binom{n}{4}}
\end{align*}
and the claim follows.

Comment:


*

*In (1) we change the order of summation of the left-most sum by exchanging $n_2\to n-3-n_2$.

*In (2) we change the order of the second sum $n_1\to n_2-n_1$.

*In (3) we observe the index $n_0=0$ can be skipped,  since it does not contribute anything. This does also hold for the other indices.

*In (4) we do a small trick and write $n_0$ as summation of $n_0$ times $1$.

*In (5) we write the index region more conveniently.  We observe the number of summands given by the index range
\begin{align*}
1\leq  n_{-1}\leq n_0\leq n_1\leq n_2\leq n-3
\end{align*}
is  the number of ordered $5$-tuples $(n_{-1},n_0,n_1,n_2,n_3)$ between $1$ and $n-3$.  This number is given by the binomial coefficient in (6).

We are  now well prepared to calculate OP's sum. We obtain for integral $n\geq r\geq 2$:
  \begin{align*}
&\color{blue}{\sum_{n_{r-2}=0}^{n+1-r}\left(\sum_{n_{r-3}=0}^{n+1-r-n_{r-2}}\left(\sum_{n_{r-4}=0}^{n+1-r-\left(n_{r-3}+n_{r-2}\right)}\cdots\left(\sum_{n_{1}=0}^{n+1-r-\sum_{k=2}^{r-2}n_{k}}\left(\sum_{n_{0}=0}^{n+1-r-\sum_{k=1}^{r-2}n_{k}}n_{0}\right)\right)\right)\right)}\\
&\qquad=\sum_{n_{r-2}=0}^{n+1-r}\sum_{n_{r-3}=0}^{n_{r-2}}\sum_{n_{r-4}=0}^{n_{r-3}}\cdots\sum_{n_{1}=0}^{n_{2}}\sum_{n_{0}=0}^{n_1}n_{0}\\
&\qquad=\sum_{n_{r-2}=1}^{n+1-r}\sum_{n_{r-3}=1}^{n_{r-2}}\sum_{n_{r-4}=1}^{n_{r-3}}\cdots\sum_{n_{1}=1}^{n_{2}}\sum_{n_{0}=1}^{n_1}\sum_{n_{-1}=1}^{n_0}1\\
&\qquad=\sum_{1\leq n_{-1}\leq n_0\leq \cdots\leq n_0\leq n_{-1}\leq n+r-1}1\\
&\qquad=\binom{r+(n+1-r)-1}{r}\\
&\qquad\,\,\color{blue}{=\binom{n}{r}}
\end{align*}
and   the claim follows.

