Let $A \in M_n(\mathbb{R})$ be a square matrix such that each $a_{ij}$ (with $i^\text{th}$ row and $j^\text{th}$ column) are represented by $$a_{ij}=\sum_{k=i}^nk^j$$
Prove or disprove that for every $n\in\mathbb{Z}>0$ that $\det(A)=n!\cdot(n-1)! \cdots 2!$
$$\ $$
I initially tried to prove it via induction, where we assume that the case is true for $A \in M_{n-1}(\mathbb{R})$ and then aim to show that $\det(B)=\ $ $n!*\det(A)$ for $B \in M_{n}(\mathbb{R})$, but I didn't get too far with that. My second attempt (which was probably what I should have started with) was simple determinant rules, where after some manipulation I got $$\det(A)=n! \det \left(\begin{matrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 2^2 & \cdots & 2^{n-1}\\ 1 & 3 & 3^2 & \cdots & 3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & n & n^2 & \cdots & n^{n-1} \end{matrix}\right)$$ and again I'm not sure where to go on from here. I appreciate any help, even it's a nudge towards the right direction.
Thank you kindly!
As Crostul says the matrix you have is a Vandermonde Matrix. This matrix has a special property for solving the determinants, but first let's define $\alpha$.
$\alpha$ are the values of the second column of the matrix, as we see you evaluate $\alpha$ as $\{1,2,3 \dots n\}$. So for solving the determinant of this matrix we have: $\prod_{1\leq i < j}{(\alpha_{j}-\alpha_{i})}$
If you expand this expression then you will end up with the equivalent $(n-1)!(n-2)!(n-n+1)!$. Let's call this Vandermonde Matrix $B$.
You were looking for a proof that:
$Det(A) = n!.Det(B) = n!\cdot \prod_{1\leq i < j}{(\alpha_{j}-\alpha_{i})} = n!(n-1)!(n-2!)\dots (n-n+1)!$
So you now have obtained the remaining part of what you were looking getting the proof. Hope you understand it.