Determinant of $n \times n$ matrix whose entries are given by $m_{ij} = i^{n-j}$

Question

Let $M_{n}$ denote the $n \times n$ matrix whose entries are given by $m_{ij} = i^{n - j}$, I believe that we have that $\det(M_{1}) = 1$ and $\det(M_{n}) = (-1)^{n-1}\cdot(n-1)! \cdot \det(M_{n-1})$ for $n \geq 2$. I have been trying to prove that this statement via induction on $n$, but I am unsure how to proceed with the induction step. If we assume that this is true for all $M_{k}$ up to some $k \in \mathbb{N}$ and consider $M_{k+1}$ then certainly $M_{k}$ is a submatrix of $M_{k+1}$ but I am unsure how to use that information to my advantage as my linear algebra is a bit rusty.

Answer 1

Here is your matrix: \begin{equation} A = \begin{pmatrix}1 & 1 & \cdots & 1\\ 2^{n-1} & 2^{n-2} & \cdots & 1\\ \cdots & \cdots & \cdots & \cdots\\ n^{n-1} & n^{n-2} & \cdots & 1 \end{pmatrix} \end{equation}

Here is the Vandermonde version of this matrix: \begin{equation} B = \begin{pmatrix}1 & 1 & \cdots & 1\\ 1 & \cdots & 2^{n-2} & 2^{n-1} &\\ \cdots & \cdots & \cdots & \cdots\\ 1 & \cdots & n^{n-2} & n^{n-1} \end{pmatrix} \end{equation}

It is well-known (and you can find plenty of proofs online) that $$ \det(B) = \prod_{k=1}^{n-1} k!$$

Since your columns are reversed, we get $$ \det(B) = (-1)^{\frac{n(n-1)}{2}}\prod_{k=1}^{n-1} k!$$

EDIT: The Vandermonde matrix has determinant $$ \prod_{1 \mathop \le i \mathop < k \mathop \le n} \left({k - i}\right)$$ which is the same as what I wrote.