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?

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

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