Let $F$ be a field, $V$ be a $F$-vector space of dimension $n$ and $L^k(V,F)$ be the space of $k$-multilinear forms $f:V^k\rightarrow F$.
I was reading Henri Cartan's Differential Form, where he defined the subspace $A^k(V,F)$ of $k$-alternating forms of $L^k(V,F)$ to be the forms $f(x_1,\cdots,x_k)$ such that if $x_i=x_{i+1}$ for some $i$, then $f=0$.
He then proved that if $char (F) \not= 2$, then a multilinear form is alternating:
Let $f\in L^k(V,F)$ and suppose $x_i=x_{i+1}$ for some $i$, let $\sigma = (i \ \ i+1)\in S_n$, then $f(x_{\sigma(1)},\cdots ,x_{\sigma (i)}, x_{\sigma (i+1)}, \cdots ,x_k)=f(x_1,\cdots ,x_k)$ since $x_i=x_{i+1}$ by assumption.
On the other hand, $f(x_{\sigma(1)},\cdots ,x_{\sigma (i)}, x_{\sigma (i+1)}, \cdots ,x_k)=sgn (\sigma) f(x_1,\cdots,x_k)=-f(x_1\cdots,x_k)$.
Hence we have $f(x_1\cdots, x_k)=-f(x_1\cdots, x_k)$, or $2f(x_1\cdots, x_k)=0$. Now since $char (F)\not= 2$, we can divide both sides by $2$ and get $f=0$, hence $f\in A^k(V,F)$ and $L^k(V,F)=A^k(V,F)$.
But doesn't that contradict the face that $\dim (L^k(V,F))=n^k$ and $\dim (A^k(V,F))=\binom{n}{k}$?
How did you deduce that $L^k(V,F)=A^k(V,F)$? What you proved was that there is an isomorphism between $A^k(V,F)$ and a certain substpace of $L^k(V,F)$.