Consecutive Matrices Clarification $$
\begin{vmatrix}
1 & 2 & 3 & \cdots & n \\
n+1 & n+2 & n+3 & \cdots & 2n \\
2n+1 & 2n+2 & 2n+3 & \cdots & 3n \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
n^2-n+1 & n^2-n+2 & n^2-n+3 & \cdots & n^2
\end{vmatrix}
$$
My instructor gave me this matrix in this form to find the determinant of it. My question is not to how find the determinant, but what or how to figure out the the fourth row down? Or what is the pattern of the fourth row?

Is this clearer?
$$
\begin{vmatrix}
1 & 2 & 3 & \cdots & n-2 & n-1 & n \\
n+1 & n+2 & n+3 & \cdots & 2n-2 & 2n- 1 &2n \\
2n+1 & 2n+2 & 2n+3 & \cdots& 3n-2 & 3n- 1 & 3n \\
\vdots & \vdots & \vdots & \vdots & \vdots& \vdots& \vdots \\
n^2-n+1 & n^2-n+2& n^2-n+3 & \cdots & n^2-2 & n^2-1 & n^2
\end{vmatrix}
$$
The pattern for each entry is $a_{ij}= (i-1)n+j$.
