I am trying to understand the proof of Engel's theorem and I am stuck with the following step. Assuming we have an invariant element $v$ in $V$ a vector space carrying a representation of $L$, where $L$ acts by nilpotent operators. We want to find a basis of $V$ with respect to which every element of $L$ is upper triangular. This can be assumed to exist when $V$ has dimension $n$, then we look at the dim$(V) = n+1$ case. We quotient $V$ by the submodule $\mathbb{K}v$ and to get an $n$-dimensional representation of $L$, allowing us to apply the inductive hypothesis and conclude an $b_i$ upper triangular basis for $V/\mathbb{K}v$.
So far so good. But now one "lifts the basis" up to $V$, takes the union with $v$ to get a new basis $$ v,b'_1, \dots, b'_{n} $$ and bingo we are done! However I do not understand what this lifting process actually means. Sure we can choice of element $b'_i$ in the preimage of each $b_i$ and taking the union with $v$ gives a basis of $V$, but why is clear that $L$ acts on $b'_i$ to give an element in the span of $b'_{i+1}, \dots, b'_n$? I don't see why the lift still has the upper triangular property . . . I guess nilpotency comes into play here somewhere, bt I cannot see how . . .
Write the new basis as $b'_1, \dots, b'_{n}, \color{magenta}{v}$. Then you see that $L$ acts on $b_{i}'$ to give an element of $b_{i+1}', \dots ,b_{n}', \color{magenta}{v}$.