I feel like I've got the answer, but I've never been good at putting what I think into words.

$\begin{vmatrix}
n\_{11} & n\_{12} \\
n\_{21} & n\_{22}
\end{vmatrix} = 0 = n\_{11}n\_{22} - n\_{12}n\_{21}$

$\begin{vmatrix}
n\_{11} + 1 & n\_{12} \\
n\_{21} & n\_{22} + 1
\end{vmatrix} = n\_{11}n\_{22} + n\_{11} + n\_{22} + 1 - n\_{12}n\_{21} = \begin{vmatrix}
n\_{11} & n\_{12} \\
n\_{21} & n\_{22}
\end{vmatrix} + n\_{11} + n\_{22} + 1 = n\_{11} + n\_{22} + 1$

Which shows that it's true for a 2\*2 matrix.  

Looking at other 2\*2 matrices

$\begin{vmatrix}
n\_{11} + 1 & n\_{12} \\
n\_{21} & n\_{22}
\end{vmatrix} = n\_{11}n\_{22} + n\_{22} - n\_{12}n\_{21} = \begin{vmatrix}
n\_{11} & n\_{12} \\
n\_{21} & n\_{22}
\end{vmatrix} + n\_{22} = n\_{22}$

Similarly:  

$\begin{vmatrix}
n\_{11} & n\_{12} \\
n\_{21} & n\_{22} + 1
\end{vmatrix} = n\_{11}$

$\begin{vmatrix}
n\_{11} & n\_{12} + 1 \\
n\_{21} & n\_{22}
\end{vmatrix} = -n\_{21}$

$\begin{vmatrix}
n\_{11} & n\_{12} \\
n\_{21} + 1 & n\_{22}
\end{vmatrix} = -n\_{12}$

Positive if on the main diagonal, negative if on the side diagonal.  

Using that, we can show that  

$\begin{vmatrix}
n\_{11} + 1 & n\_{12} & n\_{13} \\
n\_{21} & n\_{22} + 1 & n\_{23} \\
n\_{31} & n\_{32} & n\_{33} + 1
\end{vmatrix} = (n\_{11} + 1)(n\_{22} + n\_{33} + 1) - n\_{12}n\_{21} - n\_{13}n\_{31} = n\_{11}n\_{22} - n\_{12}n\_{21} + n\_{11}n\_{33} - n\_{13}n\_{31} + n\_{11} + n\_{22} + n\_{33} + 1 = n\_{11} + n\_{22} + n\_{33} + 1$

And in a similar fashion we can apply this to higher order matrices.  

But I haven't the slightest clue how to word or show this "similar fashion".