linear algebra – 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:

  1. Add the i-th with the (i-1)-th;
  2. Add the last row to all, getting two determinants, one of which is 0 (because of the (n-1)-st column);
  3. 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?