Proof that field of constants is trivial Let $d$ be $k$-derivation of $L=k(x_1,x_2,\dots)$ (field of rational functions over field k) defined by $d(x_i)=x_{i+1}$ for $i=1,2,\dots$. Then $L^d=k$.
Book says it's easy to prove it but i dont have any idea.

It is necessary for the characteristic of $k$ to be zero, for if $\mathrm{char}(k)=p$ then $d(x_1^p)=0$.
So lets assume $\mathrm{char}(k)=0$ and $f(x_1,\dots,x_n)\in L\backslash k$ (by definition: $L=\cup_n k(x_1,\dots,x_n)$). So $n\geq 1$ and we can assume further that $f\not\in k(x_1,\dots,x_{n-1})$. Then we have:
$$d(f) = \frac{\partial f}{\partial x_1}d(x_1)+\cdots+\frac{\partial f}{\partial x_n}d(x_n) \\ = \frac{\partial f}{\partial x_1}x_2+\cdots+\frac{\partial f}{\partial x_n}x_{n+1}. $$ 
But $f$ is a non-constant element of $(k(x_1,\dots,x_{n-1}))(x_n)$ so $\frac{\partial f}{\partial x_n}\neq 0,$ (here we use $\mathrm{char}(k)=0$) and so $d(f) = A + Bx_{n+1}$ for $A,B\in k(x_1,\dots,x_n), B\neq 0.$ So $d(f)$ is a nonzero element of $(k(x_1,\dots,x_n))(x_{n+1}) = k(x_1,\dots,x_{n+1})$. 
