I come to you as a humble pilgrim of mathematics with yet another basic question... I thank you for your patience!
So, lets say that $L$ is a nilpotent Lie algebra over field $F$. I'm trying to show that the killing form of $L$ is identically zero....
I mean, Cartans first critereon says that if $L$ is solvable $\iff$ $k(x,y)=0$ for all $x \in L$ and $y \in L'$. I feel like we are almost there!! Insights appreciated, thanks!
Since $L$ is a nilpotent Lie algebra, then (assuming that $\dim L<\infty$) there is some basis $B$ of $L$ such that the matrix associated to $\operatorname{ad}(X)$ with respect to $B$ is a strictly superior triangular matrix, for any $X\in L$ (it's a consequence of Engel's theorem). So, if $X,Y\in L$,$$k(X,Y)=\operatorname{tr}\bigl(\operatorname{ad}(X)\circ\operatorname{ad}(Y)\bigr)=0.$$