How to find incremental n-determinant with all ones on main diagonal and (almost) similar rows? So, the question is in the title.
$$\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
1 & 1 & 3 & 4 & \cdots & n - 1 & n \\
1 & 2 & 1 & 4 & \cdots & n - 1 & n \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}$$
This determinant equals
$$(-1)^{n}\cdot(n - 1)!$$
But how can you find it?
 A: As mentioned by nejimban in the comments, we can make the matrix upper triangular and use the fact that the determinant is invariant under row operations. However, it should be $(-1)^{n-1}(n-1)!$ instead of $(-1)^n(n-1)!$
$$\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
1 & 1 & 3 & 4 & \cdots & n - 1 & n \\
1 & 2 & 1 & 4 & \cdots & n - 1 & n \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}\!\begin{aligned}
\ \overset{r_2-r_1}{\longrightarrow} \
\end{aligned}\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
0 & -1 & 0 & 0 & \cdots & 0 & 0 \\
1 & 2 & 1 & 4 & \cdots & n - 1 & n \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}\!\begin{aligned}
\ \overset{r_3-r_1}{\longrightarrow} \
\end{aligned}$$

$$\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
0 & -1 & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & -2 & 0 & \cdots & 0 & 0 \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}\!\begin{aligned}
\ \overset{}{\longrightarrow} \
\end{aligned}\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
0 & -1 & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & -2 & 0 & \cdots & 0 & 0 \\
0 & 0 & 0 & -3 & \cdots & 0 & 0 \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
0 & 0 & 0 & 0 & \cdots & 2-n & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 & 1-n \\
\end{vmatrix}\!\begin{aligned}
\  \
\end{aligned}$$

and since the determinant of a triangular matrix is the product of its diagonal elements we have

$$\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
1 & 1 & 3 & 4 & \cdots & n - 1 & n \\
1 & 2 & 1 & 4 & \cdots & n - 1 & n \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}$$
$$=1\cdot(-1)\cdot(-2)\cdot(-3)\cdot\cdot\cdot(2-n)(1-n)=(-1)^{n\color{red}{-1}}\cdot(n-1)!$$
