Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $\mathbb{R}$ and $\phi:\mathfrak{g}\to\mathfrak{g}$ be a Lie algebra automorphism. 

Viewing $\mathfrak{g}$ as a linear space and $\phi$ a linear automorphism, we can say $\phi$ is *hyperbolic* if the eigenvalues of $\phi$ are disjoint from $\lbrace z\in\mathbb{C}:|z|=1\rbrace$.

Then Proposition 3.6 in Smale's paper ([here][1]) says that:

 - Suppose that $\phi:\mathfrak{g}\to\mathfrak{g}$ is a Lie algebra automorphism which is hyperbolic as a linear map. Then $\mathfrak{g}$ must be nilpotent.

He also mentioned the following result in (Exercise in Bourbaki with hints: Algebras de Lie, Ex. 21b, p. 124.):

 - Let $\mathfrak{g}$ be a finite dimensional Lie algebra having an automorphism $\phi$, 
no eigenvalue of which is a root of unity, then $\mathfrak{g}$ is nilpotent.

Do you have ideas how to prove these results? 

Thanks!

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After Vladimir Dotsenko:

$$(\phi-\lambda\gamma)[u,v]=[\phi u,\phi v]-[\lambda u,\gamma
v]=[(\phi-\lambda)u,\phi v]+[\lambda u,(\phi-\gamma) v].$$

Applying above to the pair
$\hat{u}=\lambda^i\phi^j(\phi-\lambda)^{a}u$ and
$\hat{v}=\gamma^k\phi^l(\phi-\gamma)^{b}v$ we have
$$(\phi-\lambda\gamma)[\lambda^i\phi^j(\phi-\lambda)^{a}u,\gamma^k\phi^l(\phi-\gamma)^bv]=
[(\phi-\lambda)\hat{u},\phi \hat{v}]+[\lambda \hat{u},(\phi-\gamma)
\hat{v}]$$
$$=[\lambda^i\phi^j(\phi-\lambda)^{a+1}u,\gamma^k\phi^{l+1}(\phi-\gamma)^bv]
+[\lambda^{i+1}\phi^j(\phi-\lambda)^au,
\gamma^k\phi^l(\phi-\gamma)^{b+1}v].$$

Tracing the indices we get
$$(i,j,a;k,l,b)\overset{\phi-\lambda\gamma}{\to}(i,j,a+1;k,l+1,b)\cup
(i+1,j,a;k,l,b+1),$$ and in particular
$(a,b)\overset{\phi-\lambda\gamma}{\to}(a+1;b)\cup (a;b+1)$. Then
$$(\phi-\lambda\gamma)^{m+n}[u,v]
=\sum_{a+b=m+n}[\lambda^i\phi^j(\phi-\lambda)^{a}u,\gamma^k\phi^l(\phi-\gamma)^bv]=0$$
since either $a\ge m$ or $b\ge n$.

  [1]: http://projecteuclid.org/euclid.bams/1183529092