The problem is stated as this:
"Prove that if $$G:=G_1\times G_2 \times \dots\times G_s \cong H_1 \times H_2 \times \dots\times H_t=:H,$$ where each $G_i$ and $H_i$ is finite simple group, then $s=t$ and there exists $\sigma\in S_t$ such that $G_j \cong H_{\sigma(j)}$ for $j=1,\dots,t$."
This comes after first being introduced to concept of solvable groups and composition series. We are given the Jordan-Hölder Theorem as follows:
Any two composition series for a finite group have the same length. Further, there exists a one-to-one correspondence between composition factors of the two composition series under which corresponding composition factors are isomorphic.
As far as I understand, each $G_i$ being simple implies that the only composition series for $G$ is $$G \trianglelefteq e_G$$ where $e_G=(e_{G_1},\dots,e_{G_s})$, and similarly for $H$. Is this correct? If it is, I am unsure how to apply the Jordan-Hölder Theorem in this instance, since $G$ and $H$ are merely isomorphic.
Can anyone offer any hints or a solution? Thank you for help.
HINT: $$G_1 \le G_1 \times G_2 \le G_1 \times G_2 \times G_3 \le \dots \le G_1 \times \cdots \times G_n$$ is a composition series for $G$. You can see this noting that $G_1 \times \cdots \times G_{k-1} $ is an epimorphic image of $G_1 \times \cdots \times G_k$ simply by the projection, and the kernel is $1 \times \cdots \times 1 \times G_k \cong G_k$.
Now, do the same thing with the $H_i$s and use Jordan-Hölder.