Determinant of parametric function and $0!1!2!...n!$

Question

As answer to this question, I trued to calculate the wronskian of:

$$\left| \begin{array}{ccc} e^x & e^{2x} & ... & e^{nx}\\ e^x & 2e^{2x} & ...& ne^{nx} \\ e^x & 4e^{2x} &... & n^2e^{nx}\\ ...&...&...&...\\ e^x&2^{n-1}e^{2x} & ... & n^{n-1}e^{nx}\end{array} \right| $$ What led to: $$\left| \begin{array}{ccc} 1 & 1 & ... & 1\\ 1 & 2 & ...& n \\ 1 & 4 &... & n^2\\ ...&...&...&...\\ 1 &2^{n-1} & ... & n^{n-1}\end{array} \right|$$ So I tried to calculate it for small values of n. The results were:

• $n=1\to1=0!$
• $n=2\to1=0!1!$
• $n=3\to2=0!1!2!$
• $n=4\to12=0!1!2!3!$
• $n=5\to288=0!1!2!3!4!$

So I have two questions:

1. Is it true for the general case?, I mean: $$\left| \begin{array}{ccc} 1 & 1 & ... & 1\\ 1 & 2 & ...& n \\ 1 & 4 &... & n^2\\ ...&...&...&...\\ 1 &2^{n-1} & ... & n^{n-1}\end{array} \right|=0!*1!*2!*...*(n-1)!$$

2. Is there some more closed form for this expression?

What I tried is $$0!1!2!...n!=(1)*(1*2)*(1*2*3)*...(1*2*3*...*n)=1^n*2^{n-1}*3^{n-2}*...*(n-1)^2*n$$ But I don't know how to continue.

Answer 1

Your matrix $M$ is a Vandermonde matrix. (Or at least the transpose of one with respect to the definition on wikipedia). You immediately obtain

$$\det(M)=\prod_{1\leq i<j\leq n} (\alpha_j-\alpha_i),$$ where $(\alpha_1,\alpha_2,\ldots,\alpha_n)=(1,2,\ldots,n)$, and your formula follows.

By the way, these are superfactorials and more information can be found at OEIS A000178.