In the proof of the lemma in this paper, the author makes the following claim: Fix integer $N$ and let $n \leq N$. Then the solutions $X_0,\ldots,X_N$ to the equations $$\sum_{k=0}^N X_{k} (-2^k)^n = 1 \qquad (n = 0,...,N)$$
are given by $$X_k := \prod_{j \in \{0,\ldots,N\} \backslash\{k\}}\frac{1+2^j}{2^j - 2^k}$$ for $k = 0,\ldots,N$.
Could I get some hints on how to show that this is true? The author claims that this follows from Cramer's rule and Vandermonde's expansion, but I'm not familiar with the statements he is referring to.
EDIT: I've made some progress: the problem above is equivalent to finding numbers $X_1,\ldots,X_n$ such that the following equation is satisfied: $$ \underbrace{\begin{pmatrix} 1 & 1 & \cdots & 1 \\ -2^0 & -2^1 & \cdots & -2^N \\ (-2^0)^2 & (-2^1)^2 & \cdots & (-2^N)^2 \\ \vdots & \vdots &\ddots& \vdots\\(-2^0)^N & (-2^1)^N & \cdots & (-2^N)^N \\ \end{pmatrix}}_{=: A}\begin{pmatrix} X_0 \\ X_1 \\ X_2 \\ \vdots \\ X_N\end{pmatrix} =\begin{pmatrix} 1 \\1 \\ 1 \\ \vdots \\ 1\end{pmatrix} $$
The matrix $A$ is a square Vandermonde matrix, and so $$\det(A) = \prod_{0 \leq i < j \leq N} (2^i - 2^j).$$ Thus, $A$ is invertible. Cramer's rule tells us that for $k = 0,\ldots, N$, $$X_k = \frac{\det(A_k)}{ \det(A)},$$ where $A_k$ is the matrix $A$ with the (zero-based) $k$th column replaced with $(1\; 1 \; \cdots \; 1)^\top$.
However, I'm not sure how to calculate $\det(A_k)$ and simplify the expression for $X_k$. Could I get some suggestions on how to finish the computation?
The general Vandermonde determinant $$V(x_0, x_1,\dots,x_N):= \begin{vmatrix} 1&1&1&\dots &1\\ x_0&x_1& x_2&\dots&x_N\\ x_0^2&x_1^2&x_2^2&\dots&x_N^2\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ x_0^N& x_1^N& x_2^N&\dots & x_N^N \end{vmatrix}$$ is well-known to be given by $$V(x_0,x_1,\dots,x_N)=\prod_{1\le i<j\le N}(x_j-x_i).$$ By Cramer's formula, $$X_k=\frac{\det A_k}{\det A},$$ where $$\det A=V(-2^0, -2^1,\dots,-2^N)=\prod_{1\le i<j\le N}(2^i-2^j)$$ and $$\det A_k=V(-2^0,\dots,-2^{k-1},1, -2^{k+1},\dots,-2^N)\\=\prod_{1\le i<j\le N\atop i,j\ne k}(2^i-2^j)\prod_{1\le i<k}(2^i+1)\prod_{k<j\le N}(-1-2^j).$$ Therefore, $$X_k=\prod_{1\le i<k}\frac{2^i+1}{2^i-2^k}\prod_{k<j\le N}\frac{-1-2^j}{2^k-2^j}= \prod_{j\in\{0,\dots,N\} \backslash\{k\}}\frac{1+2^j}{2^j-2^k}.$$