ad-nilpotent degree of a nilpotent Lie Algebra Let $\mathfrak{g}$ be a Lie Algebra (finite dimensional, over $\mathbb{C}$). Engel's theorem tells us that if there exists a $m\in \mathbb{N}$ such that $\mathrm{ad}(x)^m = 0$, $\forall x\in \mathfrak{g}$, then $\mathfrak{g}$ is nilpotent. And if $\mathfrak{g}$ is $(k-1)$-step nilpotent (i.e. the $k$-th term of the lower central series of $\mathfrak{g}$ is the first one that is 0, or equivalently $\mathrm{ad}(x_1)\mathrm{ad}(x_2) \ldots \mathrm{ad}(x_k) = 0$ $\forall x_1, \ldots, x_k \in \mathfrak{g}$), it is clear that $$\min \big\{m\in \mathbb{N} : \mathrm{ad}(x)^m = 0 \forall x\in \mathfrak{g} \big\} \leq k.$$
Can we find an example where the previous inequality is not an equality?
If this is a very basic fact in the theory, I apologize. 
Edit: I've been asked to share de Graaf's example, here it is.
 It is a 17-dimensional (nilpotent) Lie algebra over $\mathbb{Q}$ that is 3-Engel and of nilpotency class 4.
(The table has to be read as follows: $\langle 2,4,17|-3\rangle$ means that $[x_2,x_4] = -3 x_{17}$. If there are more tuples starting with $\langle 2,4$ then
one has to take the sum, so if there also was $\langle 2,4,13|-2\rangle$ then
$[x_2,x_4] = -2 x_{13} - 3 x_{17}$.)
$$[ \langle2, 4, 17| -3\rangle,\; \langle2, 6, 3| -3\rangle,\; \langle2, 7, 1| 1\rangle,\; \langle2, 9, 1| 2\rangle,\; \langle2, 12,
9| 1\rangle,\; \langle2, 13, 10| 1\rangle,\; \langle2, 14, 11| 1\rangle,\; \langle2, 15, 13| 1\rangle,\;
\langle2, 16, 14| 1\rangle,\; \langle5, 16, 17| -3\rangle,\; \langle7, 15, 17| 1\rangle,\; \langle7, 16, 3| -2\rangle,\; \langle8,
15, 3| 3\rangle,\; \langle9, 15, 17| -1\rangle,\; \langle9, 16, 3| -1\rangle,\; \langle10, 16, 1|
3\rangle,\; \langle11, 15, 1| -3\rangle,\; \langle12, 13, 17| 4\rangle,\; \langle12, 14, 3| 4\rangle,\; \langle12, 15, 4| -1\rangle,\;
\langle12, 16, 6| -1\rangle,\; \langle13, 14, 1| -4\rangle,\; \langle13, 15, 5| -1\rangle,\;
\langle13, 16, 7| -1\rangle,\; \langle14, 15, 7| -1\rangle,\; \langle14, 15, 9| 1\rangle,\; \langle14, 16, 8| -1\rangle,\;
\langle15, 16, 12| -1\rangle ]$$
 A: A Lie algebra $\mathfrak{g}$ satisfying the equation $ad(x)^m = 0$ for all $x \in \mathfrak{g}$ is called an Engel-$m$ Lie algebra. One of the key steps in Zelmanov's solution of the restricted Burnside problem is that any finitely generated Engel-$m$ Lie algebra is nilpotent. Zelmanov won the Fields medal for this work.
Let $E(t,m)$ be the free Lie algebra on $t$-generators subject to the Engel-$m$ identity. Then it follows that $E(t,m)$ is nilpotent and therefore finite dimensional, although its dimension and class depend on the characteristic of the ground field $k$. 
Let us assume that $char(k) = 0$ for simplicity. Then it is not difficult to show that $E(t,2)$ is nilpotent of class $2$, and $E(2,3)$ is nilpotent of class $3$. The dimension of $E(t,2)$ is $t + \binom{t}{2}$, and the dimension of $E(2,3)$ is $5$.
But already $E(3,3)$ is actually nilpotent of class $4$, so this gives an example of a Lie algebra where the required inequality is strict. 
More information can be found in the works of Michael Vaughan-Lee and Gunnar Traustason --- see, for example, http://people.bath.ac.uk/gt223/paper01.pdf. Traustason observes at the bottom of page 12 of this paper that "it is easy to construct an Engel-$3$ Lie algebra of class $4$".
I don't know the dimension of $E(3,3)$, but I'm sure that Willem de Graaf does. This paper contains more relevant information.
A: I computed the example of Traustason (see his remark). We start with the  free-nilpotent Lie algebra 
of class 4 with 3 generators $x_1$, $x_2$, $x_3$. It has dimension 32. Then we divide out the ideal generated by
all brackets containing $x_1$ three times, or containing $x_2$, $x_3$ at least two times. The quotient is
an Engel-3-Lie algebra of nilpotency class 4 and of dimension 11. 
It is easy to write down explicit Lie brackets. This should be an example of least possible dimension.
However, the next case, to find an Engel-4-Lie algebra of nilpotency class 7, of least possible
dimension is more complicated, and I am frightened to do the calculation.
