Verifying that the determinant is equal to $1!2!3!...(n-1)!$ Verifying that the determinant is equal to $1!2!3!...(n-1)!$
$$|A|=
\begin{vmatrix}
1 &1 & \dots &1\\
1 &2 & \dots &2^{n-1}\\
1 &3 & \dots &3^{n-1}\\
& & \dots\\
1 &n & \dots &n^{n-1}\\
\end{vmatrix}=1!2!3!...(n-1)!
$$
I used the definition of a determinan with Minors and Cofactors, ie.
$$|A|=\sum_{j=1}^{n-1}i^{j-1}\cdot A_{ij}=i^{1-1=0}\cdot A_{i 1}+i^{2-1=1}\cdot A_{i 2}+i^{3-1=2}\cdot A_{i 3}+\dots+i^{n-1}\cdot A_{i n-1}$$
So, what we want to prove is equal to:
$$|A|=\sum_{j=1}^{n-1}i^{j-1}\cdot A_{ij}=1!2!3! \dots (n-1)!$$
Proving for $n=1$
$$|A|=\sum_{j=1}^{n-1}i^{j-1}\cdot A_{ij}=|1|=0!=1$$
INDUCTIVE H:
Asume
$$|A|=\sum_{j=1}^{n-1}i^{j-1}\cdot A_{ij}=i^{0}\cdot A_{i 1}+i^{1}\cdot A_{i 2}+i^{2}\cdot A_{i 3}+\dots+i^{n-1}\cdot A_{i n-1}=1!2!3!...(n-1)$$
So, for $n+1$ we have:
$$|A|=\sum_{j=1}^{n}i^{j-1}\cdot A_{ij}=i^{0}\cdot A_{i 1}+i^{1}\cdot A_{i 2}+i^{2}\cdot A_{i 3}+\dots+i^{n-1}\cdot A_{i n-1}+i^{n}\cdot A_{i n}$$
INDUCTIVE STEP:
$$|A|=\sum_{j=1}^{n}i^{j-1}\cdot A_{ij}=(1!2!3! \dots (n-1)!)+i^{n}\cdot A_{i n}$$
AND AT THIS POINT, I DONT KNOW WHAT TO DO.
 A: This is a special case of the Vandermonde determinant:
https://en.wikipedia.org/wiki/Vandermonde_matrix
Your determinant is
$$\prod_{1 \le i < j \le n} (j - i) = \prod_{i=1}^{n-1} \prod_{j=i+1}^n (j-i) = \prod_{i=1}^{n-1} (n-i)! = 1! \dots (n-1)!$$
A: We can evaluate this determinant as follows.
Let $A_n$ be the $n \times n$ matrix in question.
First, for each $j = n-1, n-2, \ldots, 1$, subtract $n$ times column $j$ from column $j+1$.
This results in the following:
\begin{align*}
|A_n| &=
\begin{vmatrix}
1 & 1-n & 1-n & \ldots & 1-n & \ldots & 1-n \\
1 & 2 - n & 2^2 - 2n & \ldots & 2^{j-1} - 2^{j-2}n & \ldots & 2^{n-1} - 2^{n-2}n \\
1 & 3 - n & 3^2 - 3n & \ldots & 3^{j-1} - 3^{j-2}n & \ldots & 3^{n-1} - 3^{n-2}n \\
\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & i - n & i^2 - in & \ldots & i^{j-1} - i^{j-2}n & \ldots & i^{n-1} - i^{n-2}n \\
\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & n - n & n^2 - n^2 & \ldots & n^{j-1} - n^{j-1} & \ldots & n^{n-1} - n^{n-1} \end{vmatrix} \\
&\; \\ &\; \\&\; \\ &\; \\ %vertical space
&=
\begin{vmatrix}
1 & 1-n & 1-n & \ldots & 1-n & \ldots & 1-n \\
1 & 2 - n & 2(2-n) & \ldots & 2^{j-2}(2-n) & \ldots & 2^{n-2}(2-n) \\
1 & 2 - n & 3(3-n) & \ldots & 3^{j-2}(3-n) & \ldots & 3^{n-2}(3-n) \\
\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & i - n & i(i-n) & \ldots & i^{j-2}(i-n) & \ldots & i^{n-2}(i-n) \\
\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & (-1)  & (n-1)(-1) & \ldots & (n-1)^{j-2}(-1) & \ldots & (n-1)^{n-2}(-1) \\
1 & 0 & 0 & \ldots & 0 & \ldots & 0 \\
\end{vmatrix}
\end{align*}
By cofactor expansion along the bottom row,
\begin{align*}
&= (-1)^{n+1}
\begin{vmatrix}
1-n & 1-n & \ldots & 1-n & \ldots & 1-n \\
2 - n & 2(2-n) & \ldots & 2^{j-1}(2-n) & \ldots & 2^{n-2}(2-n) \\
2 - n & 3(3-n) & \ldots & 3^{j-1}(3-n) & \ldots & 3^{n-2}(3-n) \\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
i - n & i(i-n) & \ldots & i^{j-1}(i-n) & \ldots & i^{n-2}(i-n) \\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
(-1)  & (n-1)(-1) & \ldots & (n-1)^{j-1}(-1) & \ldots & (n-1)^{n-2}(-1) \\
\end{vmatrix} \\
&\; \\ &\; \\&\; \\ &\; \\ %vertical space
&= (-1)^{n+1} (1-n)(2-n)(3-n)\cdots (-2)(-1) \\
&\; \\ %vertical space
& \quad \quad \quad \quad \quad \cdot \quad 
\begin{vmatrix}
1 & 1 & \ldots & 1 & \ldots & 1 \\
1 & 2 & \ldots & 2^{j-1} & \ldots & 2^{n-2} \\
1 & 3 & \ldots & 3^{j-1} & \ldots & 3^{n-2} \\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & i & \ldots & i^{j-1} & \ldots & i^{n-2} \\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1  & (n-1) & \ldots & (n-1)^{j-1} & \ldots & (n-1)^{n-2} \\
\end{vmatrix} \\
&\; \\ %vertical space
&= (n-1)! |A_{n-1}|
\end{align*}
By induction on $n$ since $|A_1| = 1$, it follows that
$$
|A_n| = (n-1)!(n-2)!(n-3)!\cdots 2! 1!
$$
as required.
