Pattern with the the tetration of summations. While dealing with a question with finding an explicit form for a sequence I noticed something:
$$\sum_{x_0=0}^{n-1} 1=\frac{n}{1!}$$
$$\sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} 1=\frac{n(n-1)}{2!}$$
$$\sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} \sum_{x_2=0}^{x_1-1} 1= \frac{n(n-1)(n-2)}{3!}$$
$$\sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} \sum_{x_2=0}^{x_1-1} \sum_{x_3=0}^{x_2-1} 1= \frac{n(n-1)(n-2)(n-3)}{4!}$$
...

Question:
How can I prove the pattern I am seeing?

I'm thinking about induction in which case the base case is proved. But then assuming $p(n)$ is true I need to show $p(n+1)$ is true. But how can I write $p(n)$ as notation is troubling me, and how can I go from there.
 A: It's a cascasded form of the summation along a diagonal of Pascal's triangle, otherwise known as the hockey stick identity, i.e. 
$$\sum_{r=0}^m \binom ra=\binom {m+1}{a+1}$$
Assume $p$ levels of cascaded summations. Working from inside out we have 
$$\begin{align}
&\quad \sum_{x_0=0}^{n-1}
\sum_{x_1=0}^{x_0-1}
\sum_{x_2=0}^{x_1-1}\cdots 
\sum_{x_{p-3}=0}^{x_{p-4}-1}
\sum_{x_{p-2}=0}^{x_{p-3}-1}
\sum_{x_{p-1}=0}^{x_{p-2}-1}
\quad \quad 1\\
&=
\sum_{x_0=0}^{n-1}
\sum_{x_1=0}^{x_0-1}
\sum_{x_2=0}^{x_1-1}\cdots
\sum_{x_{p-3}=0}^{x_{p-4}-1} 
\sum_{x_{p-2}=0}^{x_{p-3}-1}
\underbrace{\sum_{x_{p-1}=0}^{x_{p-2}-1}
\binom{x_{p-1}}0}\\
&=
\sum_{x_0=0}^{n-1}
\sum_{x_1=0}^{x_0-1}
\sum_{x_2=0}^{x_1-1}\cdots
\sum_{x_{p-3}=0}^{x_{p-4}-1} 
\underbrace{\sum_{x_{p-2}=0}^{x_{p-3}-1}
\quad\binom{x_{p-2}}1}\\
&=
\sum_{x_0=0}^{n-1}
\sum_{x_1=0}^{x_0-1}
\sum_{x_2=0}^{x_1-1}\cdots
\underbrace{\sum_{x_{p-3}=0}^{x_{p-4}-1} 
\quad \binom{x_{p-3}}2}\\\\
&=\qquad\vdots\\\\
&=
\sum_{x_0=0}^{n-1}
\sum_{x_1=0}^{x_0-1}
\underbrace{\sum_{x_2=0}^{x_1-1}
\binom{x_2}{p-3}}\\
&=
\sum_{x_0=0}^{n-1}
\underbrace{\sum_{x_1=0}^{x_0-1}
\;\;\;\binom{x_1}{p-2}}\\
&=
\underbrace{\sum_{x_0=0}^{n-1}
\quad\binom{x_0}{p-1}}\\
&=
\qquad\binom{n}{p}
\end{align}$$
i.e. 
$$\color{red}{\boxed{\sum_{x_0=0}^{n-1}\sum_{x_1=0}^{x_0-1}\sum_{x_2=0}^{x_1-1}
\cdots \sum_{x_{p-1}=0}^{x_{p-2}-1}1=\binom np}}$$
Putting $p=4$ gives the final equation in your question, i.e. 
$$\begin{align}
&\quad \sum_{x_0=0}^{n-1}\sum_{x_1=0}^{x_0-1}\sum_{x_2=0}^{x_1-1}
\sum_{x_{3}=0}^{x_{2}-1}1\\
&=\sum_{x_0=0}^{n-1}\sum_{x_1=0}^{x_0-1}\sum_{x_2=0}^{x_1-1}\binom{x_2}1\\
&=\sum_{x_0=0}^{n-1}\sum_{x_1=0}^{x_0-1}\binom{x_1}2\\
&=\sum_{x_0=0}^{n-1}\binom{x_0}3\\
&=\binom n4\\
&=\frac {n(n-1)(n-2)(n-3)}{4!}\end{align}$$
A: I will change your notation a bit and use $x_{-1}\equiv n$ instead.
Hint: for $m>0$,
$$
\sum_{x_{0}=0}^{x_{-1}-1}\cdots\sum_{x_{m}=0}^{x_{m-1}-1}1=\sum_{x_{0}=0}^{x_{-1}-1}\left[\sum_{x_{1}=0}^{x_{0}-1}\cdots\sum_{x_{m}=0}^{x_{m-1}-1}1\right]=\cdots
$$
You should use an induction hypothesis on the inner sum at this point.
A: Note we can develop a combinatorial argument which gives total number of ways as ${n\choose r}$ where r is total number of summations
