Isomorphism of field extension 
Let $F$ be a field. I need to prove that if $\sigma$ is an isomorphism of $F(\alpha_1,...,\alpha_n)$ with itself such that $\sigma|_F = id_F$ and $\sigma(\alpha_i)=\alpha_i$ for $i=1,...n$, then $\sigma$ is the identity.

I've tried many things but nothing worked. The main problem is that I do not have a nice description of the elements in $F(\alpha_1,...,\alpha_n)$... Any help?

An element of that field looks as $\;f(\alpha_1,...,\alpha_n)\;$ , with $\;f(x_1,...,x_n)\in F(x_1,...,x_n)\;$
Thus, we have that
$$f(\alpha_1,...,\alpha_n)=\sum_{i_1,...,i_n}a_{i_1,...,i_n}\alpha_1^{i_1}\cdot\ldots\cdot\alpha_n^{i_n}\implies$$
$$\sigma f(\alpha_1,...,\alpha_n)=\sum_{i_1,...,i_n}\sigma a_{i_1,...,i_n}\;\sigma\alpha_1^{i_1}\cdot\ldots\cdot\sigma\alpha_n^{i_n}=\sum_{i_1,...,i_n}a_{i_1,...,i_n}\alpha_1^{i_1}\cdot\ldots\cdot\alpha_n^{i_n}=f(\alpha_1,...,\alpha_n)$$
