Suppose we have an $n \times n$ matrix and we know there are exactly $N$ zero entries. The determinant is the sum of $n!$ terms, each formed by selecting exactly one entry from each row and column and multiplying them. If a given term comes from selecting one or more zeroes then that term becomes zero. This raises the question of how many of the terms can be nonzero.
Intuitively I imagine the maximum numbers are achieved if we take a matrix of all $1$s and start putting in zeroes, starting with entries $(1,2), (1,3), \ldots, (1,n)$ so the only nonzero entry in that row is at $(1,1)$, and then moving to the next row and putting zeroes in $(2,1), (2,3), \ldots, (2,n)$ so the only nonzero entry in that row is at $(2,2)$, and so on until we're filling in zeroes on the last row at $(n,1), (n,2), \ldots, (n,n-1)$. At this stage we have the identity matrix and there is exactly one nonzero term.
I imagine this has already been proved somewhere but cannot find a good reference. Could someone please provide a proof, or better yet, a book where this is proved as part of some wider theory? Ideally by something more elegant than induction?