As already mentioned, we can enlarge the field, hence let us assume it to be algebraically closed and not algebraic over a finite field. Then in this context, for finite-dimensional algebras (Lie or not) $\mathfrak{g}$, the following are equivalent:
- there exists an automorphism with no root of unity as eigenvalue
- there exists a diagonalizable automorphism with no root of unity as eigenvalue
- there exists an algebra grading in a torsion-free abelian group with $\mathfrak{g}_0=0$
- there exists an algebra grading in $\mathbf{Z}$ with $\mathfrak{g}_0=0$
One goes from (1) to (2) by picking the diagonalizable part in the Jordan decomposition. (2) yields (3) since the eigenspace decomposition provides a grading in the quotient of $K^*$ by its subgroup of roots of unity. (3) yields (4) by picking the subgroup generated by the degrees, and picking a suitable homomorphism to $\mathbf{Z}$, and finally (4) yields (1) by picking a non-root of unity $t$ and define an automorphism to be multiplication by $t^n$ on $\mathfrak{g}_n$.
Note that all we use of $\mathfrak{g}$ is that its automorphism group is Zariski-closed in $\mathrm{GL}(\mathfrak{g})$; that $\mathfrak{g}$ is Lie plays no role.
We next show that (4) implies nilpotent; this works in a considerably broader context.
If $\mathfrak{g}$ is an arbitrary Lie algebra (over an arbitrary commutative ring), graded in $\mathbf{Z}\smallsetminus\{0\}$ and with bounded degree (i.e., $\mathfrak{g}_n=0$ for $|n|$ large), then $\mathfrak{g}$ is nilpotent.
More precisely, if $\mathfrak{g}$ is graded in $\{-n,\dots,-1,1,\dots,n\}$, then it is $5n$-nilpotent (possibly this is not optimal; the filiform Lie algebras show that we cannot do better than $\sim 2n$).
Let us prove this in the broader realm of Leibniz algebras. An algebra is Leibniz if it satisfies the Leibniz identity $(xy)z-(xz)y=x(yz)$ for all $x,y,z$. Note that a Lie algebra is a Leibniz algebra satisfying in addition $xx=0$ for all $x$.
First immediate fact: in a Leibniz algebra, every product of $n$ elements is a linear combination of $n$-fold right products, i.e., of the form $((\cdots ((x_1x_2)x_3)\dots )x_n)$.
Write $R_x(y)=yx$. Then $R_xR_y=R_yR_x+R_{yx}$ by the Leibniz identity. Thus a Leibniz algebra $\mathfrak{g}$ is $k$-nilpotent if and only if $R_{x_1}\dots R_{x_k}=0$ for all $x_1,\dots,x_k\in\mathfrak{g}$.
Now suppose that $\mathfrak{g}$ is graded in $\mathbf{Z}\smallsetminus\{0\}$ and we have $x_1,\dots,x_k$, with $x_i\in\mathfrak{g}_{n_i}$, with all $|n_i|\le n$. Consider $v=R_{x_1}\dots R_{x_k}$. We call the set of $v$ obtained this way by $r(n,k)$.
Whenever $n_{i+1}>0>n_i$, we can write $R_{x_i}R_{x_{i+1}}=R_{x_{i+1}}R_{x_i}+R_{x_{i+1}x_i}$, noting that $x_{i+1}x_i$ has degree $n_i+n_{i+1}$ and $|n_i+n_{i+1}|\le n-1$. We reiterate the process with the new term involving $R_{x_{i+1}}R_{x_i}$, so as to eventually obtain $v=R_{x_{\sigma(1)}}\dots R_{x_{\sigma(k)}}+w$, where $w$ is a sum of elements in $r(n-1,k-1)$ and, for some $j$, all $n_{\sigma(i)}$ are $>0$ for $i\le j$ and $<0$ for $i>j$. Write $r^+(n,k)$ the set elements in $r(n,k)$ with this additional condition. Let $R(n,k)$ and $R^+(n,k)$ be the set of sums of elements in $r(n,k)$, resp $r^+(n,k)$.
Then we have proved $R(n,k)\subset R^+(n,k)+R(n-1,k-1)$. By induction, this yields
$$R(n,k)\subset R^+(n,k)+R^+(n-1,k-1)+\dots,R^+(1,k-n+1)$$
thus
$$R(n,5n)\subset R^+(n,5n)+R^+(n-1,5n-1)+\dots,R^+(1,4n+1)$$
Now assume that $\mathfrak{g}$ is graded in $\{-n,\dots,n\}\smallsetminus\{0\}$. Then the maximal difference between two degrees is $2n$. It follows that every element in $R^+(n,k)$ for $k\ge 4n+1$ is zero: indeed, such an element either involves $2n+1$ consecutive $R_{x_i}$ for $n_i>0$, or $2n+1$ consecutive $R_{x_i}$ for $n_i<0$, which shifts the degree by $\ge 2n+1$ and hence is zero. It follows that $R(n,5n)=0$.
By linearity, this implies that $R_{x_1}\dots R_{x_{5n}}=0$ for all $x_i$, and by the Leibniz identity this implies that $\mathfrak{g}$ is $5n$-nilpotent.
Initial answer(for finite-dimensional Lie algebras in characteristic zero only): The statement (as in Bourbaki) is equivalent to: every (finite dimensional) Lie algebra with an invertible derivationself-derivation is nilpotent.
Since every derivationself-derivation of a semisimple Lie algebra is inner (this is elementary, see e.g. http://amathew.wordpress.com/2010/01/30/derivations-of-semisimple-lie-algebras-and-the-abstract-jordan-decomposition/), you already know the Lie algebra $\mathfrak{g}$ is solvable. Now from the derivation you can define a semidirect product $\mathfrak{h}=\mathfrak{g}\rtimes\mathfrak{a}$, where $\mathfrak{a}$ is the one-dimensional Lie algebra. Since the derivation is invertible, the derived subalgebra $\mathfrak{h}'$ is equal to $\mathfrak{g}$. Since the derived subalgebra of any solvable Lie algebra is nilpotent (consequence of Ado's Theorem [Edit: no it's not needed, see below]*), it follows that $\mathfrak{g}$ is nilpotent.
[Edit:(*) consider the adjoint representation of any solvable Lie algebra \mathfrak{h}; triangulate it (over an algebraic closure), so that $\mathfrak{h}'$ is mapped to algebra of upper nilpotent matrices. Since the adjoint representation has central kernel, this shows that $\mathfrak{h}'$ is nilpotent. Also as pointed out by Jim
It remains the question whether more generally, I assume characteristic zero in the whole argumentarbitrary characteristic, a finite-dimensional Lie algebra admitting an invertible self-derivation, is always nilpotent.]
