Let $N_i \subseteq G_i$ for $i = 1, ..., n$ be Normal subgroups. Show that $\prod_{i=1}^{n}N_i \subseteq \prod_{i=1}^{n}G_i$ is a Normal subgroup a) Let $N_i \subseteq G_i$ for $i = 1, ..., n$ be Normal subgroups. Show that $\prod_{i=1}^{n}N_i \subseteq \prod_{i=1}^{n}G_i$ 
 is  a Normal subgroup
b)  Find an isomorphism $\prod_{i=1}^{n}G_i / \prod_{i=1}^{n}N_i\cong\prod_{i=1}^{n}(G_i/N_i)$  
I know that the product of two Normal subgroups is Normal but I'm not sure if I need this for a) and after some time of thinking about b) I still have  no clue. 
I've tried for a few hours and it's getting frustrating, there must be some kind of elegant solution.. Help is appreciated!
EDIT: $\prod_{i=1}^{n}G_i$ means $G_1 \bigotimes G_2 \bigotimes  ... \bigotimes G_n$ 
 A: b) Hint: consider the homomorphism 
$$\phi:\prod_{i=1}^{n} G_i \to \prod_{i=1}^n G_i / N_i$$
Defined by:
$$\phi(g_1,\dots,g_n)=(g_1 N_1,\dots,g_n N_n)$$
And use the first Isomorphism Theorem.
Proving a) can be done immediately after  proving that $N_1 \times N_2$ is a normal subgroup of $G_1 \times G_2$ (i.e use induction).
A: (a) Let $\textbf{g}\in \displaystyle\bigotimes_{i=1}^k G_i$. Then $\textbf{g}=(g_1,g_2,\ldots,g_k)$. Let $\textbf{n}\in  \displaystyle\bigotimes_{i=1}^k N_i$. Then $\textbf{n}=(n_1,n_2,\ldots,n_k)$. To prove that $ \displaystyle\bigotimes_{i=1}^k N_i$ is normal in $ \displaystyle\bigotimes_{i=1}^k G_i$ you need to show that, $$\textbf{gn}\textbf{g}^{-1}\in  \displaystyle\bigotimes_{i=1}^k N_i$$Now observe that, \begin{align}\textbf{gn}\textbf{g}^{-1}&=(g_1,g_2,\ldots,g_k)(n_1,n_2,\ldots,n_k)(g_1,g_2,\ldots,g_k)^{-1}\\&=(g_1,g_2,\ldots,g_k)(n_1,n_2,\ldots,n_k)(g_1^{-1},g_2^{-1},\ldots,g_k^{-1})\\&=(g_1n_1g_1^{-1},g_2n_2g_2^{-1},\ldots,g_kn_kg_k^{-1})\\&\in  \displaystyle\bigotimes_{i=1}^k N_i\quad(\color{red}{?})\end{align}Hence $ \displaystyle\bigotimes_{i=1}^k N_i$is normal in $ \displaystyle\bigotimes_{i=1}^k G_i$.
(b) Define the homomorphism $$\varphi: \displaystyle\bigotimes_{i=1}^k G_i\to \displaystyle\bigotimes_{i=1}^k \dfrac{G_i}{N_i}$$ by $\varphi(g_1,g_2,\ldots,g_k)=(g_1N_1,g_2N_2,\ldots,g_kN_k)$. If the function is indeed an homomorphism then you can apply the First Isomorphism Theorem to conclude that, $$\ker \varphi=\left\{\textbf{g}\in  \displaystyle\bigotimes_{i=1}^k G_i\mid \varphi(\textbf{g})=(N_1,N_2,\ldots , N_k)\right\}= \displaystyle\bigotimes_{i=1}^k N_i$$but first of all prove that the function $\varphi$ is indeed an homomorphism. Can you do that?
