A simple representation theoretic explanation of this Linear Algebra determinant formula Let $A: \mathbb{R}^n \to \mathbb{R}^n$ be an $\mathbb{R}$-linear map. If $1 \leq k \leq n$, we consider
$\bigwedge^k A: \bigwedge^k \mathbb{R}^n \to \bigwedge^k \mathbb{R}^n $.
What is a simple representation-theoretic argument that
$\det( \bigwedge^k A ) = \det(A)^\binom{n-1}{k-1}$ ?
This is I think a formula by Sylvester, if I remember correctly. Can we say perhaps that the map $A \mapsto \det( \bigwedge^k A )$ is $GL(n)$-invariant (and maybe something else) so it must be a power of $\det(A)$?
I know it is ``elementary'', but I would like the precise proposition please. 
 A: First of all we can clearly focus on invertible matrices (if $A$ is not invertible, $\bigwedge^k A$ isn't either for $k$ within the given bounds, so the determinant is $0$). 
Then, for these notice that $f: A\mapsto \det (\bigwedge^k A)$ is a continuous group morphism $GL_n(\mathbb{R})\to \mathbb{R}^\times$, in particular since $GL_n(\mathbb{R})' = SL_n(\mathbb{R})$ and $\det$ realizes an (topological group) isomorphism $GL_n(\mathbb{R})/SL_n(\mathbb{R}) \to \mathbb{R}^\times$, it follows that $f$ factors through the determinant. 
So $f = g\circ \det$ for some continuous morphism $g : \mathbb{R}^\times \to\mathbb{R}^\times$. But one can easily check that the only such continuous morphisms are of the form $x\mapsto \epsilon(x)^i |x|^l$ for some $i\in \{0,1\}$ and $l$ a real number, where $\epsilon(x)$ denotes the sign of $x$, i.e. $+/- 1$.
Therefore $f(A) = \epsilon(\det(A))^i |\det(A)|^l$ for some $i,l$ as above. Now check on some well-chosen matrices to find what $i$ and $l$ are (if $f(A)<0$ for some $A$, $\epsilon = 1$, otherwise $\epsilon = 0$, and then $l$ should follow by checking on a homotethy and getting a suitable power of the factor)
I don't know if this is "representation-theoretic"
