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Because $\theta$ is a Lie algebra automorphism, for every $y$,Conjugation therefore also proves that $\mathrm{ad}(\theta x)$ is a nilpotent operator, and sends every power of $\mathrm{ad}x$ to the corresponding power of $\mathrm{ad}(\theta x)$. Applying this to the defining finite exponential of a nilpotent Lie algebra derivation yields

Write $D=\mathrm{ad}x$, and choose $N$ with $D^N=0$. The exponential of a nilpotent Lie algebra derivation is the finite linear operatorFor a derivation of a Lie algebra, induction on $k$ using the Leibniz rule and Pascal's identity provesExpanding $[Ey,Ez]$ and grouping by $k=j+l$ consequently givesTerms 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 coefficientso $\exp (-D)$ is the inverse. We have proved bracket preservation and invertibility, hence $\exp (\mathrm{ad}x)$ is a Lie algebra automorphism.