Find the determinant of nth order $$
    \begin{vmatrix}
    2 & 2 & ... & 2 & 2 & 1 \\
    2 & 2 & ... & 2 & 2 & 2 \\
    2 & 2 & ... & 3 & 2 & 2 \\
    ... & ... & ... & ... & ... & ... \\
    2 & n-1 & ... & 2 & 2 & 2 \\
    n & 2 & ... & 2 & 2 & 2
    \end{vmatrix}
$$
I got this in my linear algebra homework. In the task, it is required to find the determinant of a matrix by the method of representing the sum of determinants. By that I mean this property of determinants:
$$
    \begin{vmatrix}
    a & b+e \\
    c & d+f \\
    \end{vmatrix} = \begin{vmatrix}
    a & b \\
    c & d \\
    \end{vmatrix} + \begin{vmatrix}
    a & e \\
    c & f \\
    \end{vmatrix}
$$
What I have tried:

*

*Add the i-th with the (i-1)-th;

*Add the last row to all, getting two determinants, one of which is 0 (because of the (n-1)-st column);

*Tried to get n on the diagonal, this is what I got:
$$
    \begin{vmatrix}
    2 & 3 & ... & n-1 & n & n \\
    2 & 3 & ... & n-1 & n & n+1 \\
    2 & 3 & ... & n & n & n+1 \\
    ... & ... & ... & ... & ... & ... \\
    2 & n & ... & n-1 & n & n+1 \\
    n & 3 & ... & n-1 & n & n+1
    \end{vmatrix}
$$
The closest one to mine from StackExchange was this one. But I didn't manage to link these two determinants.

No matter how I transform it, nothing worked for me. Any ideas?
 A: Let's do some elementary operations on rows and columns:
$$\begin{align}
\begin{vmatrix}
    2 & 2 & ... & 2 & 2 & 1 \\
    2 & 2 & ... & 2 & 2 & 2 \\
    2 & 2 & ... & 3 & 2 & 2 \\
    ... & ... & ... & ... & ... & ... \\
    2 & n-1 & ... & 2 & 2 & 2 \\
    n & 2 & ... & 2 & 2 & 2
    \end{vmatrix} 
&\overset{1}= 
\begin{vmatrix}
    2 & 2 & ... & 2 & \color{red}2\downarrow & 1 \\
    0 & 0 & ... & 0 & 0 & 1 \\
    0 & 0 & ... & 1 & 0 & 1 \\
    ... & ... & ... & ... & ... & ... \\
    0 & n-3 & ... & 0 & 0 & 1 \\
    n-2 & 0 & ... & 0 & 0 & 1
    \end{vmatrix} \\[2mm]
&\overset{2}= 2\begin{vmatrix}
    0 & 0 & ... & 0 & \leftarrow\color{red}1 \\
    0 & 0 & ... & 1 & 1 \\
    ... & ... & ... ... & ... & ... \\
    0 & n-3 & ...  & 0 & 1 \\
    n-2 & 0 & ...  & 0 & 1
    \end{vmatrix} \\[2mm]
&\overset{3}= (-1)^{n-1}2\begin{vmatrix}
    0 & 0 & ... & 1  \\
    ... & ... & ... ... & ...  \\
    0 & n-3 & ...  & 0  \\
    n-2 & 0 & ...  & 0 
    \end{vmatrix} \\[2mm]
&\overset{4}= (-1)^{n-1 + \frac{(n-2)(n-3)}{2}}2\cdot (n-2)!
\end{align}$$
Explanation:
$1)$ Subtract the first row from all others
$2)$ Expand through $(n-1)$-th column
$3)$ Expand through the first row
$4)$ Calculate the anti-diagonal determinant
