Prove that $G=S_{1}\times S_{2}\times\cdots\times S_{k} $ 
Let $G$ be a finite group and let $N_1,N_2,\ldots,N_k$ be normal subgroups. Suppose that $\bigcap_{i=1}^{k}N_{i}=\{1\}$ and that $G/N_i=S_i$ is simple. If all the groups $S_i$ are non-isomorphic prove that $G=S_{1}\times S_{2}\times\cdots\times S_{k}$.

Help me some hints.
 A: I mentioned earlier that induction was a good hint but received little response so I feel the best thing to do is write things out properly.
So lets use induction on $k$.
For the case when $k=1$, we have $N_1=1$ and so $S_1 \equiv G/N_1 \equiv G$.
Suppose by induction we have the result for all naturals less than or equal to $k$, and consider $k+1$. Now, let $N= \bigcap_{i=1}^k S_i$. We have $N \cap N_{k+1}=1$ and so $N \cdot N_{k+1}$ is a subgroup of $G$ isomorphic to $N \times N_{k+1}$. If $gh \in N \cdot N_{k+1}$ and $c \in G$, then $cghc^{-1}=(cgc^{-1})(chc^{-1}) \in N \cdot N_{k+1}$ so that $N \cdot N_{k+1}$ is also normal. Now, $N \cdot N_{k+1}$ contains $N_{k+1}$ and $G/N_{k+1}$ is simple so $N \cdot N_{k+1}$ must either be $N_{k+1}$ or $G$. 
Suppose it is $N_{k+1}$. Then $N \subset N_{k+1}$ and so $\bigcap_{i=1}^k N_i=\bigcap_{i=1}^{k+1} N_i=1$. So by inductive assumption, we can conclude that $G \equiv S_1 \times ... \times S_k$. Notice that this gives a good choice of composition series for $G$. In particular, we take $1 \subset S_1 \subset S_1 \times S_2 \subset ... \subset S_1 \times S_2 \times... \times S_k=G$ and we observe the quotients are $S_i$ for $1 \leq i \leq k$. Now, we can also make another composition series by taking $N_{k+1}$ as the maximal subgroup and continuing from there (by finiteness, this series will terminate). But now we have a contradiction because one of the quotients in this series is $S_{k+1}$ but this does not appear in the other series and by Jordan-Holder, these should be the same.
Thus, $N \cdot N_{k+1}=G$ and so $G/N_{k+1} \equiv N$. What this means is that for each $g \in S_{k+1}$, we can find some $h \in N$ so that $h \cong g \mod N_{k+1}$. Moreover, we see that if we take the map $T: G \to S_1 \times ... \times S_{k+1}$ given by projection onto each coordinate, then $h \mapsto (1,1,...,g)$.
Notice that our choosing $k+1$ here as opposed to any other $i$ for $1 \leq i \leq k+1$ was arbitrary. In other words, for each $N_i$ and $g_i \in N_i$, we can produce $h_i$ so that $h_i \mapsto (1,1,...,g_i,...,1)$. This shows that the map in question is a surjection as desired.
