Show that $f\in A_{n-1}(V)$ or $f\in A_n(V)$ is decomposable.
$f\in A_k(V)$ is decomposable if there exists a $a_1,...,a_k\in V^\wedge$ such that $f=a_1\wedge...\wedge a_k$
In this case "let $n=dim(V)$" and the -1 follows.
I have seen many pages from this site before and I know that it is almost required of me to show what I have done, unfortunately in this case I have done nothing, I am not sure how to approach this at all, it is my first question on decomposing tensors.
I do however know it to be true (http://mathworld.wolfram.com/Decomposable.html is the most recent tab) but I'm on page 10 of search-engine results, so I finally registered.
Thanks
One case is trivial, since $A_n V$ is a one dimensional space, every vector in it is of the form $a e_1\wedge\cdots\wedge e_n$, which is decomposable.
I've been thinking about the other case and I think you could prove it using the isomorphism between $A_1$ and $A_{n-1}$ (choose a base for $V = A_1 V$ and define a linear function in the natural way for the basis), but I'm not sure yet. I'll edit as soon as I know the right answer.
EDIT1: Consider a basis $\{e_i\}$ of $V$, then we can define an isomorphism between $A_1 V$ and $A_{n-1}V$ by $f(e_i)=e_1\wedge\cdots\wedge \hat{e}_i\wedge\cdots\wedge e_n$ (the hat means you remove that element from the product), and extending it linearly (this is equivalent to using the Hodge star operator suggested in the link provided in the comments).
Then, taking its inverse, $g$, and considering an arbitrary element of $\eta\in A_{n-1}V$, $$ \eta = \sum_{k=1}^n a_k e_1\wedge\cdots\wedge \hat{e}_k\wedge\cdots\wedge e_n, $$ we have that $g(\eta) = \sum_{k=1}^n a_ke_k := v_1$. We can then extend $v_!$ to a whole basis of $V$, $\{v_i\}$. The next part consist of showing that $f(v_1) = K v_2\wedge\cdots\wedge v_n$ for some constant $K$, which should require some work.
Note: There are different ways to approach this subject, so if you're thinking about antisymmetric multilinear functions, whenever I talk about a basis o $V$ think about it's dual basis.