I have written down a proof for an exercise from Chapter 4.5 in Dummit & Foote's Abstract Algebra. I think there might be something wrong with my proof, since most of the solutions I found emphasizing on proving $G_I$ is a direct product, but I somehow wrote quite a lot proving $K_i \leq G_I$ instead. I would appreciate if you can help locate the mistake.
Exercise: Let $K_1, K_2, \dots, K_n$ be non-abelian simple groups and let $G = K_1 \times K_2 \times\dots\times K_n$. Prove that every normal subgroup of $G$ is of the form $G_I = K_{i_1} \times K_{i_2} \times\dots\times K_{i_m}$ for some subset $I = \{i_1, i_2, \dots, i_m\}$ of $\{1,2,\dots,n\}$.
My attempts: Suppose $H$ is a non-trivial normal subgroup of $G$. We prove first the following claim.
Claim: For each $i \in \{1,2,\dots,n\}$, either $K_i \leq H$, or every $h \in H$ can be written as $h = k_1k_2\dots k_n$ with $k_j \in K_j$ and $k_i = 1$. It suffices to establish the case $i = 1$.
Assume that there is some element $h = k_1k_2\dots k_n$ with $k_1 \neq 1$. For any $g\in K_1$, by the normality of $H$ we know $^gh = \,^gk_1k_2\dots k_n \in H$, and it follows $^ghh^{-1} = [g, k_1] \in H$. Hence, the group $[K_1,k_1] $ generated by $\{[g,k_1]:g\in K_1\}$ is a subgroup of $H\cap K_1$.
It remains to prove that $[K_1,k_1] $ is normal in $K_1$, which follows from the following fact: Given any $x \in K_1$, $^x[g,k_1] = [xg,k_1][x,k_1]^{-1} \in [K_1,k_1]$. Also, $[K_1, k_1] \neq \{1\}$, because there must be some $g$ not commutative with $k_1$, otherwise $\langle k_1 \rangle$ would be an abelian normal subgroup of $K_1$, a contradiction. Thus, $[K_1, k_1] = K_1$ by the simplicity of $K_1$, that is, $K_1 \leq H$. We have proved the claim.
Denote $J=\{j: K_j \leq H, j\in\{1,2,\dots,n\}\}$. Thus $G_J$ is a subgroup of $H$. By the claim above, every element $h\in H$ can be written as $h = k_{j_1}k_{j_2}\dots k_{j_m}$, hence also an element of $G_J$. Therefore, $G_J = H$, and we have proved the exercise.
edit: I have correct the typos as Mr. Arturo Magidin pointed out in his first and second comment, also I have replaced "Henceforth" by "Hence". I apologize for my carelessness; I am really not in form yesterday. Also I am not a native speaker of English, so I make a lot of mistakes typing a long paragraph like this when feeling low.
I also modify the notation of conjugate as suggested by Mr. Arturo Magidin. So $^xy$ stands for $xyx^{-1}$, and $[x,y]$ for $xyx^{-1}y^{-1}$. Following this convention, the equation $$^x[g,k_1] = [xg,k_1][x,k_1]^{-1}$$ means $$^x[g,k_1] = x(gk_1g^{-1}k_1^{-1})x^{-1} = xgk_1g^{-1}(x^{-1}k_1^{-1}k_1x)k_1^{-1}x^{-1} = (xgk_1g^{-1}x^{-1}k_1^{-1})(k_1xk_1^{-1}x^{-1}) = [xgk_1(xg)^{-1}k_1^{-1}](xk_1x^{-1}k_1^{-1})^{-1} = [xg,k_1][x,k_1]^{-1}.$$
