Prove this equality: $det(\mathrm I_n-\mathbf B\mathbf A)=2^n$

Question

If $\mathbf A$ and $\mathbf B$ are square matrices ($n$ dimensional) which verifies $\mathbf A\mathbf B=-\mathrm I_n$, then prove that: $$det(\mathrm I_n-\mathbf B\mathbf A)=2^n$$

I'm struggling on this problem, because I can't find a link to $2^n$. So, I need a quick hint.

Edit: Sorry for making a new post here, but I realized that my previous question had a mistake, so I beg you to help me again, because it was too simple to prove $det(\mathrm I_n-\mathbf A\mathbf B)=2^n$.

Answer 1

Note that, since $AB = -I$, $A$ is surjective (as a linear transformation), and hence injective (since we are in finite dimensions). So, $A$ is invertible. Let $C = A^{-1}$. Then $$ -I = -AC = AB $$ $$ \Rightarrow B = -C = -A^{-1} $$ Hence, $AB = BA$, and so $I-BA = I-AB=2I$, and you are done.

Answer 2

Exercise: for any given square matrices $X,\,Y$ the eigenvalues of $XY$ and $YX$ are same (multiplicity taken into account).

Therefore, the eigenvalues of $BA$ are the same as of $AB$, and all of them are equal to $-1$.

Finally, $\det$ of a matrix is a product of eigenvalues of this matrix. IT's clear to see that the eigenvalues of $I-BA$ are all equal to $1-(-1)=2$, which allows to conclude.