Set $n=[L:K]$

First, let there be $x\in L$ such that $y=x-\sigma(x)$ Then $$\operatorname{Tr}(y)=(x-\sigma(x))+(\sigma(x)-\sigma^2(x))+\cdots +(\sigma^{n-1}(x)-\sigma^n(x))=0$$ because $x=\sigma^n(x)$

Now, let $\operatorname{Tr}(y)=0$ Choose $z\in L$ with $\operatorname{Tr}(z)\neq 0$ Then there exists $x\in L$ with $$x\operatorname{Tr}(z)=y\sigma(z)+(y+\sigma(y))\sigma^2(z)+\cdots +(y+\sigma(y)+\cdots+\sigma^{n-1}(y))\sigma^{n-1}(z).$$ Since $\operatorname{Tr}(z)\in K$ we have $$\sigma(x)\operatorname{Tr}(z)= \sigma(y)\sigma^2(z)+(\sigma(y)+\sigma^2(y))\sigma^3(z)+\cdots +(\sigma(y)+\cdots +\sigma^{n-2})\sigma^{n-1}(z)+(\sigma(y)+\cdots +\sigma^{n-1}(y))\sigma^n(z).$$ Now remember that $\operatorname{Tr}(y)=0$ We obtain \begin{eqnarray*} (x-\sigma(x))\operatorname{Tr}(z)&=&y\sigma(z)+ (y+\sigma(y))\sigma^2(z)+\cdots +(y+\sigma(y)+\cdots+\sigma^{n-1}(y))\sigma^{n-1}(z)\\ &&-\sigma(y)\sigma^2(z)-(\sigma(y)+\sigma^2(y))\sigma^3(z)-\cdots -(\sigma(y)+\cdots +\sigma^{n-2})\sigma^{n-1}(z)+yz\\ &=&y\operatorname{Tr}(z), \end{eqnarray*}so $y=x-\sigma(x)$ as we wanted to show.