Dimension of vector space of all $n$-linear functions Let $F$ be a field.  I am trying to compute the dimension of the vector space of all $n$ linear functions $D:F^k \rightarrow F$
I figured this is a pretty standard calculation but the only place so far I have found help is http://en.wikipedia.org/wiki/Exterior_algebra
Question:  Is the dimension of vector space of all $n$ linear functions from $D:F^k \rightarrow F$ just $n^k$
If we consider alternating $n$-linear functions instead is the dimension $ n \choose k$?
 A: Suppose that $D:F^n\rightarrow F$ is $n$-linear. In other words,
$$D(ca_1,\ldots,a_n)=cD(a_1,\ldots,a_n)$$
$$\vdots$$
$$D(a_1,\ldots,ca_n)=cD(a_1,\ldots,a_n)$$
for all $a_1,\ldots,a_n,c\in F$, 
and
$$D(a_1+b,a_2,\ldots,a_n)=D(a_1,a_2,\ldots,a_n)+D(b,a_2,\ldots,a_n)$$
$$\vdots$$
$$D(a_1,\ldots,a_{n-1},a_n+b)=D(a_1,\ldots,a_{n-1},a_n)+D(a_1,\ldots,a_{n-1},b)$$
for all $a_1,\ldots,a_n,b\in F$.
Therefore, for all $a_1,\ldots,a_n\in F$,
$$D(a_1,\ldots,a_n)=a_1\cdots a_n D(1,\ldots,1)$$
Thus, the entire function $D$ is determined by where $(1,\ldots,1)\in F^n$ goes. We can choose to send it anywhere in $F$. Therefore, the dimension of the space of $n$-linear functions $D:F^n\rightarrow F$ is 1.

This can also be seen by noting that, for any $F$-vector spaces $V$ and $W$, the space of $n$-linear functions $V^n\rightarrow W$ is naturally isomorphic to the space of linear functions $\underbrace{V\otimes\cdots\otimes V}_{n\text{ times}}\rightarrow W$ (this is by the universal property of tensor products). In our case, we have $V=W=F$. But for any vector space $V$, we also have
$$F\otimes V\,\cong\!\!\! V$$
(for the same reason: a bilinear map $f:F\times V\rightarrow V$ satisfies $f(a,v)=af(1,v)$ for all $a\in F$, $v\in V$, and therefore is determined by its action on $V$). Therefore,
$$\underbrace{F\otimes\cdots\otimes F}_{n\text{ times}}\,\cong\!\!\! F$$
and therefore the question becomes: what is the dimension of the space of linear maps $F\rightarrow F$, which is clearly 1.
