Consider an $n \times n$ matrix $A_n$ with elements $a_{i,j}$ such that $a_{1,j},a_{2,j},a_{3,j},\ldots$ is the sequence of numbers not divisible by $j+1$ in increasing order starting from $1$ (e.g. $1,2,4,5,7,8,10,\ldots$ for $j = 2$).
For fun, I have computed the determinant for $n \le 8$ and then conjectured that:
$$\lvert A_n \rvert = (-1)^{n+1}$$
Is the conjecture true? How would you prove it?
Your conjecture is true, and I'm surprised I've never seen this problem before; it's very charming.
The last column of the matrix is always $(1,2,3,\dots, n)^T$. Subtracting this from every other column, we obtain a matrix of the following form:
$$ \begin{bmatrix} 0,...,0 & 1 \\ L_n & *\end{bmatrix}$$
where $L_n$ is lower-triangular $(n-1)\times(n-1)$ with ones on the diagonal. (We can be more explicit if we like, but there is no need to.) Cofactor expansion along the first row shows that $\det(A_n) = (-1)^{n+1}\det(L_n)$, and $\det(L_n)=1$.