Let $G$ be a finite group with no abelian subfactor in its composition series.
Is $G$ obtained from simple groups by iterating semidirect products?
(Initially it was asked whether $G$ is a direct product of simple groups, but $A_5\wr A_5$ was mentioned as an immediate counterexample.)
There are groups that look like wreath products, but where the base group has no complement, so they are not semidirect products. The theory is described in an old paper of mine (and probably elsewhere):
D. F. Holt, Embeddings of group extensions into Wreath products, Quar. J. Math. (Oxford) 29 (1978), 463--468.
It can be seen as a non-abelian analogue of Shapiro's lemma in cohomology.
Roughly speaking, let $D$ be a group extension of $N$ by $H$ (meaning $N$ is the normal subgroup), suppose that the group $G$ contains $H$ as a subgroup, and let $\Omega$ be the set of (right or left) cosets of $H$ in $G$ with the transitive (right or left) action of $G$ on $\Omega$.
Then there is corresponding wreath product like extension $E$ of $N^\Omega$ by $G$, where the factors of $N^\Omega$ are permuted under conjugation corresponding to the action of $G$ on $\Omega$, and the normalizer in $E$ of a factor of the base group modulo the other factors of the base group is naturally isomorphic to $D$.
Furthermore, the extension $E$ is non-split if and only if $D$ is.
So, for example, we could take $D = {\rm Aut}(A_6)$ (as in spin's answer) with $N = {\rm Inn}(A_6) \cong A_6$, and $H = C_2^2$, and $G = A_5$, so $|\Omega| = 15$. This results in a non-split extension of $A_6^{15}$ by $A_5$, which cannot be constructed from simple groups by taking iterated semidirect products.
The group $\operatorname{Aut}(A_6)$ has a unique simple normal subgroup, $\operatorname{Inn}(A_6) = A_6$. It is a nonsplit extension $$1 \rightarrow A_6 \rightarrow \operatorname{Aut}(A_6) \rightarrow C_2 \times C_2 \rightarrow 1.$$
