Lower bound for matrix determinant algorithm
The complexity of your algorithm on $n \times n$ matrices satisfies the recurrence
$$
T(n)=nT(n-1) + O(n^3),
$$
with a base case of $T(1) = O(1)$. In particular, $T(n) \geq n! T(1)$.
In the other direction, let $S(n) = T(n)/n!$. Then
$$
S(n) = S(n-1) + O(n^3/n!),
$$
with a base case of $S(1) = O(1)$. Since $\sum_{n=0}^\infty n^3/n!$ converges, we get $S(n) = O(1)$, and so $T(n) = O(n!)$. (The same argument also proves the lower bound.)