In the paper, *A polytope related to empirical distributions, plane trees, parking functions, and the associahedron*, [Pitman and Stanley](https://arxiv.org/pdf/math/9908029.pdf) studied the $n$-dimensional polytope
$$\Pi\_n(\mathbf x)=\{y\in\Bbb{R}^n: y\_i\geq0, y\_1+\cdots+y\_i\leq x\_1+\cdots+x\_i, \text{for all $1\leq i\leq n$}\}.$$
They noted the results
\begin{align\*}
\text{Vol}(\Pi\_n(\mathbf x))
&=\frac1{n!}\sum\_{\mathbf{k}\in \mathbf{K}\_n}\binom{n}{k\_1,\dots,k\_n}\,\,x\_1^{k\_1}\cdots x\_n^{k\_n} \\
&=\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\left(\sum\_{h=1}^ix\_h\right)^{j-i+1}\right)\_{i,j=1}^n
\end{align\*}
where $\chi$ is the *indicator function* and
$$\mathbf{K}\_n=\{\mathbf{k}\in\mathbb{Z}^n: k\_i\geq0, k\_1+\cdots+ k\_n=n, k\_1+\cdots k\_i\leq i, \text{for all $1\leq i\leq n$}\}.$$
The set $\mathbf{K}\_n$ has cardinality $\vert\mathbf{K}\_n\vert=\frac1{n+1}\binom{2n}n$. Consider instead the determinant
$$\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n.$$
A *composition* of $n$ is a finite sequence of positive integers summing to $n$. To help us calculate the latter determinant, we introduce the set of compositions of length-$n$
$$\mathcal{B}\_n=\{y\in\mathbb{Z}^n: y\_i\geq0, \text{$y$ is a composition of $n$ with $0$ suffixes padded if necessary}\}.$$
For example, $\mathcal{B}\_2=\{20, 11\}$ and $\mathcal{B}\_3=\{300, {\color{red}{210}}, 120, 111\}$. Note $\vert\mathbf{B}\_n\vert=2^{n-1}$.


**QUESTION.** Is this true?
$$\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n
=\frac1{n!}
\sum\_{\mathbf{k}\in \mathcal{B}\_n}(-1)^{\#(\mathbf{k})}\binom{n}{k\_1,\dots,k\_n} \,\,x\_1^{k\_1}\cdots x\_n^{k\_n},$$
where $\#(\mathbf{k})$ stands for the number of zeroes in $\mathbf{k}$.


**REMARK.** One may find it convenient to work with the alternative formulation
$$\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n
=\prod\_{k=1}^n\frac{k!}{(1+k)!}\cdot \det\left(\binom{1+j}{i}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n.$$
${\color{blue}{POSTSCRIPT}}$. Max Alekseyev's comment requires some adjustment in the set $\mathcal{B}\_n$. So, here is (I hope) the correct construction: $\mathbf{y}=(y\_1,\dots,y\_n)\in\mathcal{B}\_n$ iff $y\_1>0$; $y\_i\in\mathbb{Z}\_{\geq0}$ for all $i$; when $\mathbf{y}$ is read (cyclically) $y\_1\rightarrow y\_2\rightarrow\cdots\rightarrow y\_n\rightarrow y\_1$, each $y\_i\neq0$ is followed by $y\_i-1$ zeroes. Clearly, $y\_i\leq n$ for all $i$.


For example, $\mathcal{B}\_2=\{20, 11\}$ and $\mathcal{B}\_3=\{300, {\color{red}{201}}, 120, 111\}$ and $\mathbf{B}\_4=\{4000,3001,2020,2011,1300,1201,1120,1111\}$. Observe that $\vert \mathcal{B}\_n\vert=2^{n-1}$, still.



Catalan sequences vs composition sequences
Given the corrected definition of $\mathcal{B}\_n$ in the Postscriptum, the equality does hold and can be proved by induction on $n$.


For $n=1$, the equality is trivial. For $n>1$, let's expand the determinant in the left-hand size of the equality over the first column to get
$$\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n
= x\_1 \det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=2}^n - \int\_0^{x\_1} dx\_2 \det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=2}^n.$$


By the induction, the right-hand side in the expansion equals
$$\frac1{n!}
\sum\_{\mathbf{k}\in \mathcal{B}\_{n-1}}(-1)^{\#(\mathbf{k})}\binom{n}{1,k\_1,\dots,k\_{n-1}} \,\,x\_1 x\_2^{k\_1}\cdots x\_n^{k\_{n-1}} - \frac1{n!}
\sum\_{\mathbf{k}\in \mathcal{B}\_{n-1}}(-1)^{\#(\mathbf{k})}\binom{n}{k\_1+1,k\_2,\dots,k\_{n-1}} \,\,x\_1^{k\_1+1}x\_3^{k\_2}\cdots x\_n^{k\_{n-1}}$$
$$=\frac1{n!}
\sum\_{\mathbf{t}\in \mathcal{B}\_n}(-1)^{\#(\mathbf{t})}\binom{n}{t\_1,\dots,t\_n} \,\,x\_1^{t\_1} x\_2^{t\_2}\cdots x\_n^{t\_n},$$
where the two sums in the left-hand side correspond to the following two cases in the right-hand size:


* $t\_1=1$ and $t\_{i+1} = k\_i$ for $i=1,2,\dots,n-1$; and
* $t\_1=k\_1+1>1$, $t\_2=0$, and $t\_{i+1} = k\_i$ for $i=2,\dots,n-1$.


QED


