Show that $Z(G)=\prod\limits_{i=1}^n Z(N_i)$ 
Lemma 7.7 (From Isaacs's book): Suppose that $G= \prod\limits_{i=1}^n\; N_i$ . Then, $Z(G)=\prod\limits_{i=1}^n Z(N_i).$

Let $Z_i=Z(N_i)$ and $C_G(X)$ denoting the centralizer of $X$ in $G$. Since $N_i \trianglelefteq  G$ for all $i$ and $N_i \cap N_j = \{1\}$ for $i\neq j$, we get
$$ N_j \subset C_G(N_i)\subset C_G(Z_i) $$
and since also $N_i\subset C_G(Z_i)$, we have
$$ G= \prod N_j \subset C_G(Z_i)$$
and $Z_i \subset Z(G)$ for each $i$. Thus, $\prod Z_i \subset Z(G) ...$
Question: I'm trying to understand how to show that $Z_i \subset Z(G)$ from that, but I'm not seeing how. Thanks in advance for any hint.

You are making a meal of this
You don't need normal subgroups or $C_G$'s to answer this problem. I am not sure what $C_G$ even means. Your notation $N_i \cap N_j$ is also imprecise but I gather you are writing $N_1$ for $N_1 \times \prod_{i=2}^{n}\{1_i\}$ for example.
I will do half of the problem using elementary methods and leave the second half to you.
Consider the element $a= (g,1,\ldots, 1)$ for $g \in Z_1$. For any $b = (g_1,g_2,\ldots, g_n) \in G$ we have $g g_1 = g_1 g$ and so $$ab = (g g_1,g_2,\ldots, g_n) = (g_1 g ,g_2,\ldots, g_n) = ba$$
It follows $a \in Z(G)$. Likewise we see the elements $(1,,\ldots, 1, g, 1 \ldots, 1)$ are in $Z(G)$ where $g$ is in the corresponding $Z_i$. Since $Z(G)$ is a group we can multiply all these elements to get everything in $\prod_i Z_i$.
Thus we have proved $\prod_i Z_i \subset Z(G)$.
Now use similar methods to show $Z(G) \subset \prod_i Z_i $.
