Revisited - Number of $(0,1)$ $n \times n$ matrices with exactly $k$ ones and no row or column contains only zeros

Question

Long time listener, first time caller here. I am experimenting with matrices and want to know the number of valid combinations for a binary {0,1} $n \times n$ matrix based on density\sparsity. Combinatorics seems like it might provide a calculation method, but the inclusion-exclusion calculations are new to me. I have a thorough understanding of set theory and basics in combination/permutation calculations.

This post talks about a {0,1} $m \times n$ matrix that has no empty rows or columns. (Moreover, every row and column has at least one 1) Brian M. Scott concludes with an inclusion-exclusion calculation for a sum of possible valid combinations for an $m \times n$ matrix. Note: Because my matrix is square, I've replace $m$ with $n$.

I confirmed the first few cases with my own inspection and this works for $n=1, 2, 3$. $$ \sum_{k=0}^n\binom{n}k(-1)^k\left(2^{n-k}-1\right)^n\tag{1} $$

Another post talks about a similar {0,1} $m \times n$ matrix but calculates with a restriction number $k$ of 1s contained within. It also has the same condition that every row and column have at least one 1. 'bof' gives an inclusion-exclusion calculation, I have substituted $n$ for $m$ (this might be invalid?), and have confirmed upon inspection that the calculation is correct for $n=1,2,3$.

$$ \sum_{u=0}^n\sum_{v=0}^n(-1)^{u+v}\binom nu\binom nv\binom{(n-u)(n-v)}k\tag{2} $$

and 'bof' concludes with $$ \sum_{r=0}^n\sum_{s=0}^n(-1)^{2n+r+s}\binom nr\binom ns\binom{rs}k.\tag{3} $$

Here is the interesting part, summing $(3)$ from $k=0$ to $n$ should reproduce the calculation found in $(1)$.
And it does for $n=1,2,3,4$. But once $n\geq5$, the calculations differ! The following are tables of my results:

$n=1$ is trivial

$n=2$

k	0	1	2	3	4	sum	$(1)$	difference
	0	0	2	4	1	7	7	0

$n=3$

k	0	1	2	3	4	5	6	7	8	9	sum	$(1)$	difference
	0	0	0	6	45	90	78	36	9	1	265	265	0

$n=4$

k	0	1	2	3	4	5	~	12	13	14	15	16	sum	$(1)$	difference
	0	0	0	0	24	432	~	1812	560	120	16	1	41503	41503	0

$n=5$

k	0	1	2	3	4	5	~	21	22	23	24	25	sum	$(1)$	difference
	0	0	0	0	0	120	~	12650	2300	300	25	1	24997921	24997916	5

$n=6$

k	~	5	6	7	8	9	~	32	33	34	35	36	sum	$(1)$	difference
	~	0	720	43200	755100	6700500	~	58905	7140	630	36	1	57366997447	57366986518	10929

There is some discrepancy between $(1)$ and summing the parts of $(3)$ when $n>=5$.

Any insight on if I've made a mistaken substitution, I've used the calculations incorrectly, or something else missed would be immensely helpful.