Help with computing the determinates

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I need some help with this 5x5 matrix. I did the problem multiple times and end up with different answers. $$ \begin{vmatrix} 1 & 2 & 3 & 4 & 5\\ 2 & 3 & 4 & 5 & 1\\ 3 & 4 & 5 & 1 & 2\\ 4 & 5 & 1 & 2 & 3\\ 5 & 1 & 2 & 3 & 4\\ \end{vmatrix} $$ Multi. the 1st row with -2 then add to 2nd row.
Multi. the 1st row with -3 then add to 3rd row.
Multi. the 1st row with -4 then add to 4th row.
Multi. the 1st row with -4 then add to 4th row.
Multi. the 1st row with -5 then add to 5th row.

$$ \begin{vmatrix} 1 & 2 & 3 & 4 & 5\\ 0 & -1 & -2 & -3 & -9\\ 0 & -2 & -4 & -7 & -13\\ 0 & -3 & -11 & -14 & -17\\ 0 & -9 & -13 & -17 & -21\\ \end{vmatrix} $$ Remove a -1 from 2nd row. (not sure how to format a -1 outside the matrix.)
$$ \begin{vmatrix} 1 & 2 & 3 & 4 & 5\\ 0 & 1 & 2 & 3 & 9\\ 0 & -2 & -4 & -7 & -13\\ 0 & -3 & -11 & -14 & -17\\ 0 & -9 & -13 & -17 & -21\\ \end{vmatrix} $$
Multi. the 2nd row with 2 then add to 3rd row.
Multi. the 2nd row with 3 then add to 4th row.
Multi. the 2nd row with 9 then add to 5th row.
$$ \begin{vmatrix} 1 & 2 & 3 & 4 & 5\\ 0 & 1 & 2 & 3 & 9\\ 0 & 0 & 0 & -1 & 5\\ 0 & 0 & -5 & -5 & 10\\ 0 & 0 & 5 & 10 & 60\\ \end{vmatrix} $$
swap 3rd row with 4th row, and put a negative sign outside (makes the -1 to 1?).
$$ \begin{vmatrix} 1 & 2 & 3 & 4 & 5\\ 0 & 1 & 2 & 3 & 9\\ 0 & 0 & -5 & -5 & 10\\ 0 & 0 & 0 & -1 & 5\\ 0 & 0 & 5 & 10 & 60\\ \end{vmatrix} $$
Factor out a 5 in 3rd row (place a 5 outside of the matrix).
$$ \begin{vmatrix} 1 & 2 & 3 & 4 & 5\\ 0 & 1 & 2 & 3 & 9\\ 0 & 0 & -1 & -1 & 2\\ 0 & 0 & 0 & -1 & 5\\ 0 & 0 & 5 & 10 & 60\\ \end{vmatrix} $$
Multi. the 3rd row with 5 then add to 5th row.
$$ \begin{vmatrix} 1 & 2 & 3 & 4 & 5\\ 0 & 1 & 2 & 3 & 9\\ 0 & 0 & -1 & -1 & 2\\ 0 & 0 & 0 & -1 & 5\\ 0 & 0 & 0 & 5 & 70\\ \end{vmatrix} $$
Multi. the 4th row with 5 then add to 5th row.
$$ \begin{vmatrix} 1 & 2 & 3 & 4 & 5\\ 0 & 1 & 2 & 3 & 9\\ 0 & 0 & -1 & -1 & 2\\ 0 & 0 & 0 & -1 & 5\\ 0 & 0 & 0 & 0 & 95\\ \end{vmatrix} $$
Multiply the pivot position.
5(1*1*-1*-1*95) = 475
I plugged the matrix into my calculator and got 1875. Im not sure where i made a mistake. Thanks in advance.

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For the general case:

\begin{align*} \left| \begin{matrix} 1 & 2 & 3 & \dots & n \\ 2 & 3 & \dots & n & 1 \\ 3 & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots \\ n & 1 & 2 & \dots & n-1 \end{matrix}\right| &= \frac{n(1+n)}{2} \cdot \left| \begin{matrix} 1 & 2 & 3 & \dots & n \\ 1 & 3 & \dots & n & 1 \\ \vdots & \dots & \dots & \dots & \dots \\ 1 & n & 1 & \dots & n-2 \\ 1 & 1 & 2 & \dots & n-1 \end{matrix}\right| \\ &= \frac{n(1+n)}{2} \cdot \left| \begin{matrix} 1 & 2 & 3 & \dots & n \\ 0 & 1 & \dots & 1 & 1-n \\ \vdots & \dots & \dots & \dots & \dots \\ 0 & 1 & 1-n & \dots & 1 \\ 0 & 1-n & 1 & \dots & 1 \end{matrix}\right|_{n \times n} \\ &= \frac{n(1+n)}{2} \cdot \left| \begin{matrix} 1 & \dots & 1 & 1-n \\ \dots & \dots & \dots & \dots \\ 1 & 1-n & \dots & 1 \\ 1-n & 1 & \dots & 1 \end{matrix}\right|_{(n-1) \times (n-1)} \\ &= \frac{n(1+n)}{2} \cdot (-1) \cdot \left| \begin{matrix} 1 & \dots & 1 & 1-n \\ \vdots & \dots & \dots & \dots \\ 1 & 1-n & \dots & 1 \\ 1 & 1 & \dots & 1 \end{matrix}\right| \\ &= \frac{n(1+n)}{2} \cdot (-1) \cdot \left| \begin{matrix} 1 & 0 & \dots & 0 & -n \\ \vdots & \dots & \dots & \dots & \dots \\ 1 & -n & 0 & \dots & 0 \\ 1 & 0 & \dots & 0 & 0 \end{matrix}\right| \\ &= \frac{n(1+n)}{2} \cdot (-1) \cdot (-1)^{\frac{(n-1)(n-2)}{2}} \cdot (-n)^{n-2} \\ &= (-1)^{1+\frac{(n+1)(n-2)}{2}} \cdot\frac{(1+n)}{2} \cdot n^{n-1} \end{align*}