How does one show, by applying the properties of determinants (adding multiple of a column/row to another, factoring out a constant, etc.), that
$$ \begin{vmatrix} 1 & 2 & 3 & ... & n \\ 2 & 3 & 4 & ... & 1 \\ 3 & 4 & 5 & ... & 2 \\ {\vdots}& {\vdots}& {\vdots} & {\vdots} & {\vdots} \\ n & 1 & 2 & ... & n-1 \notag \end{vmatrix} = (-1)^{\frac{n(n-1)}{2}}\frac{n^n+n^{n-1}}{2} $$
When summing over any column or row, one gets a column/row of equal entries, namely $\frac{n(n+1)}{2}$. How would one proceed from there on to find the remaining terms?
This question is related, but doesn't reveal the strategy to solve the determinant, i.e. finding the formula.