= Solution
Fix $x$ with $x^N=0$ as a <matrix>. On the full <matrix> algebra define $L_x(y)=xy$ and $R_x(y)=yx$. These endomorphisms commute and each has $N$th power zero. The <Adjoint representation of a Lie algebra> is $\operatorname{ad}(x)=L_x-R_x$. The binomial theorem gives
$$
(\operatorname{ad}x)^m(y)=\sum_{j=0}^{m}(-1)^j\binom mjx^{m-j}yx^j.
$$
For $m=2N-1$, each summand has either $j\ge N$ or $m-j\ge N$, and hence vanishes. Therefore $\operatorname{ad}(x)$ is nilpotent on the full <matrix> algebra and on its <invariant subspace> $\mathfrak g$.
<Engel's theorem>, in its adjoint form, states that a finite-dimensional <Lie algebra> is nilpotent if every $\operatorname{ad}(x)$ is a <nilpotent endomorphism>. Applying it to all $x\in\mathfrak g$ proves
$$
\boxed{\mathfrak g\text{ is nilpotent}.}
$$
This is distinct from assuming that an arbitrary <representation> of a nilpotent algebra must act by nilpotent matrices, which is false.
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