= Solution
Write $D=\operatorname{ad}x$, and choose $N$ with $D^N=0$. The <exponential of a nilpotent Lie algebra derivation> is the finite <linear operator>
$$
E=\exp D=\sum_{j=0}^{N-1}\frac{D^j}{j!}.
$$
For a <derivation of a Lie algebra>, induction on $k$ using the Leibniz rule and Pascal's identity proves
$$
D^k[y,z]=\sum_{j=0}^k\binom{k}{j}[D^jy,D^{k-j}z].
$$
Expanding $[Ey,Ez]$ and grouping by $k=j+l$ consequently gives
$$
[Ey,Ez]=\sum_{k=0}^{2N-2}\frac1{k!}\sum_{j=0}^k\binom{k}{j}[D^jy,D^{k-j}z]
=\sum_{k=0}^{2N-2}\frac{D^k[y,z]}{k!}=E[y,z].
$$
Terms with an exponent at least $N$ vanish, which justifies extending each inner sum; and $D^k=0$ for $k\ge N$. Finally the finite product of the two exponentials has coefficient
$$
[D^k]\bigl(\exp D\exp(-D)\bigr)=\sum_{j=0}^k\frac{(-1)^{k-j}}{j!(k-j)!}=\frac{(1-1)^k}{k!},
$$
so $\exp(-D)$ is the inverse. We have proved bracket preservation and invertibility, hence \b[$\exp(\operatorname{ad}x)$ is a Lie algebra automorphism].
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