$G$ is characteristically simple $\iff$ there is simple $T$ such that $G \cong T\times T \times \cdots \times T$ 
Let $G$ be a characteristically simple finite group, i.e. it has no nontrivial characteristic subgroups. Prove there is some simple group $T$ such that $G \cong T \times T \times \cdots \times T$.

No idea how to start this one. I have tried to induct on the size of $G$ to no avail. Any help would be appreciated!
 A: Hints:
We may suppose $\;G\;$ has a non-trivial normal subgroup , otherwise the claim follows at once. 
Since $\;G\;$ is finite, choose a minimal non-trivial $\;N\lhd G\;$, and look at 
$$\;M:=\langle\;N^\phi\;:\;\;\phi\in\text{Aut}\,(G)\;\rangle$$
Prove now that $\;G\;$ is the direct product of some of the $\;N^\phi$'s .
Disclaimer: The only proof of the above I know is applying Zorn's Lemma on the set of $\;N^\phi$'s generating their own direct product. It seems to me weird to use this powerful weapon with a finite group, yet I cannot see right now a way out of it.
A: Let $G$ be a finite group which is characteristically simple. Let $T$ be
a minimal normal subgroup of $G$.
In order to prove this, we are going to pull out of thin air and define a set like following:
$\mathscr{D} = \{N \unlhd G \: | \: N = T_1 \times T_2 \times ... \times T_k\}$,
where each $T_i$ is a minimal normal subgroup of $G$ isomorphic to $T$.
The idea is from all of possible normal subgroups $N$ that are constructed by direct product of minimal normal subgroups isomorphic to $T$, we should be able to get the largest one that is equal to $G$. If it is so, $N = G$, then the theorem holds.
We choose the largest $N \in \mathscr{D}$. Assume $N \neq G$ and as $G$ is characteristically simple, $N$ is not characteristic in $G$. Hence there exists an automorphism $\phi$ of $G$ such that:
$\phi(N) \not\leq N$.
As $N = T_1 \times T_2 \times ... \times T_k$ is a direct product, any automorphism works on each of its component individually, thus there exists $i$ such that:
$\phi(T_i) \not\leq N$.
Because $\phi$ is an automorphism of $G$, so $\phi(T_i)$ is also a minimal subgroup of $G$ and isomorphic to $T_i$. Because both $N$ and $\phi(T_i)$ are normal subgroup of $G$, so is $N \cap \phi(T_i) \unlhd G$. But as $\phi(T_i) \not\leq N$, $N \cap \phi(T_i)$ is properly contained in $\phi(T_i)$. By minimality of $\phi(T_i)$, then $N \cap \phi(T_i) = 1$.
By recognition theorem (e.g. Dummit's Abstract Algebra page 171), if $N \cap \phi(T_i) = 1$, then $N \phi(T_i) \cong N \times \phi(T_i)$ is a group. Because both $N$ and $\phi(T_i)$ are normal subgroup of $G$, then $N \phi(T_i) \unlhd G$.
This shows that $N \phi(T_i) \in \mathscr{D}$, contradicts $N$ being the largest member of $\mathscr{D}$. In other words, our assumption $N \neq G$ cannot be true.
Therefore:
$G = N = T_1 \times T_2 \times ... \times T_k$,
where $T_i$ isomorphic to minimal normal subgroup $T$.
Note that we just assume $T$ is a minimal subgroup, hence not necessarily simple. It remains to check if $T$ is simple. Assume there exist $J \unlhd T_i$, then we could have:
$J \unlhd T_1 \times T_2 \times ... \times T_k = G$.
As $T_i$ is minimal $J$ could only be $1$ or equal to $T_i$, hence $T_i$ is simple, and $T$ accordingly.
A: Another possible proof of this fact is a bit longer but still deserves to be mentioned:
Suppose $G$ is a direct product of isomorphic copies of a finite simple group. It is quite obvious, that $G$ is characteristically simple (the inverse statement is the non-trivial one).
Now suppose, that $G$ is characteristically simple. Then, because every verbal subgroup is characteristic, then $G$ is also verbally simple. And if a finite group is verbally simple, then it is a direct products of isomorphic copies of a finite simple group (this fact is proved here: Does there exist some sort of classification of finite verbally simple groups?)
