(Just so this question has an answer.)
As fkraiem points out, a permutation matrix is a matrix with precisely one $1$ in each row and precisely one $1$ in each column, and zeroes elsewhere.
Your argument is correct as is your conclusion. Furthermore, $n\times n$ permutation matrices are in one-to-one correspondence with $S_n$, the group of permutations on $n$ elements (hence the name). The determinant of a permutation matrix is the sign of the corresponding permutation.
Choose any matrix with rank $n-1$ that does not have any of the standard unit vectors in its column space.
Added in response to the comment by alex.jordan.
Let $A$ be an $n \times n$ matrix with $rank(A) = n-1$ such that there are vectors $a, \, \tilde a$ with $Aa = 0, \tilde a^T A = 0$ that have all entries $ a_i , \tilde a_i \ne 0$. Then any matrix $B$ that differs from $A$ in exactly one entry has full rank, i.e. $\det B \ne 0$.
To prove this, consider such a $B$. After permuting rows and columns and rescaling, we may assume that $B_{1,1} = A_{1,1} + 1$.
First note that the first column of $A$ is a linear combination of columns $2, \dots, n$, since $Aa = 0$ and $a_1 \ne 0$. The column space of $B$ certainly contains the column space of $A$ and thus $rank(B) \ge n-1$.
If $rank(B) < n$, then $A$ and $B$ must therefore have the same column space. Hence the standard unit vector $e_1$ is in the column space of $A$, that is $Ac = e_1$ for some vector $c$. But then $0 = \tilde a^TAc = \tilde a e_1 = \tilde a_1$, contradicting the assumptions for the left and right null vectors of $A$.
Therefore $rank(B) > n-1$ and $\det B \ne 0$.
Best Answer
Adding or subtracting a row of a matrix from another does not change its determinant, so we may assume each column of the matrix has at most one entry that is 1.
Swapping rows of a matrix changes the sign of the determinant only; so if we perform row swaps so that the resulting matrix is diagonal, we'll have determined the determinant up to a sign.
So now we have a diagonal matrix whose diagonal entries are either 1 or 0. The determinant of this matrix must be $0$ or $1$; and hence, the determinant of the original matrix must be $0$, $1$, or $-1$.
(The $-1$ possibility can arise: start with the identity matrix and interchange the last two rows. The 0 possibility can arise: start with a matrix whose first column is all $1$'s. And, of course, the identity matrix shows that $1$ is a possible value of the determinant.)