= Solution
Use the standard split <weight-space decomposition> in <characteristic zero>, in particular the usual complex setting: $L=\bigoplus_\varphi L_\varphi$, with generalized weights for the nilpotent algebra $H$. The characteristic-zero condition prevents nonzero weight shifts from wrapping around; the unrestricted positive-characteristic version would need additional hypotheses.
For fixed $h\in H$, let $D=\operatorname{ad}h$. On a bracket of <vectors> from $L_\varphi,L_\psi$, the <derivation> identity gives
$$
(D-(\varphi+\psi)(h))[u,v]=[(D-\varphi(h))u,v]+[u,(D-\psi(h))v].
$$
Iterating it gives the binomial expansion of these two commuting actions. At power $2n-1$, each term kills at least one factor, because generalized weight <vectors> are killed by the $n$th power of the corresponding shifted operator. Therefore the bracket lies in the generalized eigenspace for $(\varphi+\psi)(h)$ for every $h$. Stabilization of kernels by dimension $n$ gives
$$
\boxed{[L_\varphi,L_\psi]\subseteq L_{\varphi+\psi},}
$$
where $L_\eta=0$ when $\eta$ is not a weight.
Nilpotence of $H$ gives $H\subseteq L_0$. If $H$ is a <Cartan subalgebra> and $L_0/H\ne0$, the action of $H$ on this quotient consists of nilpotent operators. <Engel's theorem> supplies a nonzero coset $v+H$ killed by $H$, giving $[H,v]\subseteq H$ with $v\notin H$, a contradiction to self-normalization. Conversely, if $L_0=H$ and $v$ normalizes $H$, then $[h,v]\in H$ for each $h$. Applying further powers of $\operatorname{ad}h$ kills this <vector> because its restriction to $H$ is nilpotent. Hence $v\in L_0=H$. This proves the equivalence.
The operator $\operatorname{ad}e$ sends $L_\varphi$ into $L_{\varphi+\alpha}$. Repetition follows a consecutive $\alpha$-string; after $r$ steps it must leave the weight set, because $r$ is the largest number of weights in any such string. Thus $\boxed{(\operatorname{ad}e)^r=0}$ on every <weight space> and hence on $L$.
Finally put $E=\operatorname{ad}e$ and $F=\operatorname{ad}f$. Expand $(E+F)^n$ as the sum of all words of length $n$, without commuting $E$ and $F$. Acting on $L_\gamma$, each word follows weights $\gamma+i\alpha+j\beta$ with $i+j$ increasing by one at each step. <Linear independence> of $\alpha,\beta$ in <characteristic zero> makes the successive weights distinct. A nonzero length-$n$ path would require $n+1$ different <weight spaces>, but $\dim L=n$. Every word therefore vanishes, giving
$$
\boxed{(\operatorname{ad}(e+f))^n=0.}
$$
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