Determinant of exterior power

Suppose $$A$$ is a $$n$$ times $$n$$ matrix.

What is the determinant of the $$i$$-th exterior power of $$A$$, in terms of determinant of $$A$$?

The answer is just $$\det(\Lambda^i(A))=\det(A)^m$$, where $$m=\binom{n-1}{i-1}$$. Indeed, by continuity it is enough to prove this when $$A$$ is diagonalisable, and by conjugation-invariance it suffices to prove it when $$A$$ is diagonal, and in that case it is straightforward. Alternatively, you can show that $$SL_n(\mathbb{C})$$ is the commutator subgroup of $$GL_n(\mathbb{C})$$ and thus that any polynomial homomorphism $$GL_n(\mathbb{C})\to\mathbb{C}^\times$$ is a power of the determinant. To find the relevant power, just take $$A$$ to be a multiple of the identity.