Suppose U$\subseteq R^m$, F:U$\to R^m$ is $C^{\infty}$, f$\in C^{\infty}(R^m)$. And $x_1,x_2...x_m$:U$\to$R are coordinates on U.{$e^*_1,e^*_2...e^*_m$} is the basis of $(R^m)^*$ dual to the basis {$e_1=(1,0,0...0)...e_m=(0,0,...1)$}. $v_1,v_2...v_m \in T_xU$.
Then why do we have the equation: (det(DF(x))$e^*_1{\bigwedge}e^*_2{\bigwedge}...{\bigwedge}e^*_m$)$(v_1,v_2...v_m$)=(det DF(x)$e^*_1,dx_1{\bigwedge}dx_2...{\bigwedge}dx_m$)$(v_1,v_2...v_m$)? Is $dx_i=e^*_i$? If so, could you show me why?
It seems that there are some small errors in the quesiton: The function $f$ is not used anywhere. Also, the right hand side of the equation looks strange, I guess there should be no $e_1^*$ there.
However, it seems to me that the point of your question is only the relation between the $e_i^*$ and the $dx_i$ (so also $F$ does not really play a role), and I can try to answer that: You have $U\subset R^m$, so for each $x\in U$, the tangent space to $U$ at $x$ is just $R^m$. Hence a one-form on $U$ is smooth function $U\to L(R^m,R)$. Now denote by $x_1,\dots,x_m$ the restrictions of the coordinate functions on $R^m$ to $U$ (so these are not just some coordinates on $U$, but specific ones). Then the one-form $dx_i$ (the extierior derivative of the $i$th coordinate function) coincides with the constant function $e_i^*:U\to L(R^m,R)$. This follows readily from the fact that the $i$th coordinate of $x+tv$ is $x_i+tv_i$, so $dx_i(x)(v)=v_i$.