I like Serena said:
Hey mathmari! (Smile)
What about induction?
We can include the 2x2 matrix for which it also holds true.
Did you know that we can evaluate any determinant like this:
$$\begin{vmatrix} a & b & c & d\\e & f & g & h\\i & j & k & l\\m & n & o & p \end{vmatrix}=a\,\begin{vmatrix} f & g & h\\j & k & l\\n & o & p \end{vmatrix}-b\,\begin{vmatrix} e & g & h\\i & k & l\\m & o & p \end{vmatrix}+c\,\begin{vmatrix} e & f & h\\i & j & l\\m & n & p \end{vmatrix}-d\,\begin{vmatrix} e & f & g\\i & j & k\\m & n & o \end{vmatrix}$$
(Wondering)
Base case:
For $n=2$ we have that $\begin{vmatrix} a & b \\ c & d\end{vmatrix}=ad-bc$.
Since $n+1=3$ elements are equal to $1$ we have that one element is $0$. That means that the result is either $1$ or $-1$.
For $n=3$ we have that $\begin{vmatrix} a & b & c\\ d & e & f \\ g & h & i\end{vmatrix}=a\,\begin{vmatrix} e & f \\ h & i\end{vmatrix}-b\,\begin{vmatrix} d & f \\g & i \end{vmatrix}+c\,\begin{vmatrix} d & e\\g & h \end{vmatrix}$.
Since $n+1=4$ elements are equal to $1$ we have that $9$ elements are equal to $0$.
If $a=b=c=0$ then the result is equal to $0$.
If at least one of the $a,b,c,d$ is non-zero, then we have from the previous case that the $2\times 2$ determinants will be equal to $1$ or $-1$. Therefore, the $3\times 3$ determinant will be equal to $-1$, $0$ or $1$.
Is this correct? Or could we improve the justification? (Wondering)
Inductive hypothesis:
We suppose that $n\times n$ determinant will is equal to $-1$, $0$ or $1$.
Inductive step:
The expansion of Laplace allows to reduce the computation of the $(n+1)\times (n+1)$ determinant to that of $n+1$ $(n\times n)$ determinants.
These $n+1$ determinants are equal to $-1$, $0$ or $1$.
We have $n+2$ elements that are equal to $1$ and the remaining ones are $0$.
How could we continue? I got stuck right now. (Wondering)
