Show that a field extension $L/K$ is separable iff the trace form is non-degenerate. Let $L$ be a finite field extension of $K$. I have the following question:

Show that $L/K$ is separable if and only if the bilinear trace form $\text{Tr}_{L/K}:L\times L \to K$ is non-degenerate. 

A proof or reference will be great. Thank you! 

This is a quite well-known fact, the proof can be find in every textbook on algebraic number theory but I would like to tell a proof which is different from the standard one but more elementary.
The fact that $L/K$ is separable means that for every $a\in L$ its minimal polynomial $f$ over $K$ does not have multiple roots. Since $f$ is irreducible it is equivalent to $f'\not\equiv 0$. Consider the logarifmic derivative of $f$(the formal one) $$\frac{f'}{f}=(\log(x-a_1)\dots(x-a_n))'=\frac{1}{x-a_1}+\dots+\frac{1}{x-a_n}$$ there $a_1\dots a_n$ are the roots of the $f$(possibly not lying in $L$). Now we expand this as a formal series $$\frac{1}{x}\sum\limits_{k=0}^{\infty}(\frac{a_1^k}{x^k}+\dots+\frac{a_n^k}{x^k})=\frac{1}{x}\sum\limits_{k=0}^{\infty}\frac{tr_{K(a)/K}(a^k)}{x^k}$$  so $f'\equiv 0$ iff $tr(a^k)=0$ for every $k$.
(EDIT $tr_{K(a)/K}(a^k)$ is equal to ${a_1^k}+\dots+{a_n^k}$ because the characteristic polnomial of $a^k$ is equal to $\pm(x-a_1^k)\dots(x-a_n^k)$ since it actually annihilates $a^k$ and has degree equal to the degree of the extension)
If $a$ is an inseparable element of $L$ then $tr_{K(a)/K}\equiv 0$ so $tr_{L/K}=tr_{L/K(a)}\circ tr_{K(a)/K}\equiv 0$. Conversely, if $tr_{L/K}\equiv 0$ then obviously there exists an inseparable element.
Finally, the form $Tr$ is degenerate iff the covector $tr$ is identically zero. 
