Can a Matrix with Only 1s and -1s Have a Determinant Divisible by 2^(n-1)?

[Ackbach]

N.B. There has been a correction in this problem, thanks to Opalg. It should read correctly now.

Here is this week's POTW:

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Let $A$ be an $n\times n$ matrix whose entries are only $1$ or $-1$. Show that $2^{n-1}$ divides $\det(A)$.

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[Ackbach]

Congratulations to Opalg for his correct and, may I say, very elegant and direct solution, not even requiring mathematical induction! See the spoiler for his solution:

Spoiler

Solution using elementary row operations.

Let $B$ be the matrix obtained from $A$ by subtracting the top row from every other row. Then $\det(B) = \det(A).$ The top row of $B$ will consist of $\pm1$s. Each element of every other row will be of the form $\pm1 \pm 1$, so will be equal to $0$ or $\pm2$. So we can take a factor $2$ out of each of these $n-1$ rows, leaving a matrix $C$ in which every element is $0$ or $\pm1$. Then $\det( C)$ will be an integer, and $\det(A) = \det(B) = 2^{n-1}\det( C).$ So $\det(A)$ is a multiple of $2^{n-1}.$