The Lemmas:
-Lem. 1: Let $k\in\mathbb{N}$. Let $G_{1}$, ..., $G_{k}$ be groups. Then $$Z(G_{1}\times...\times G_{k})=Z(G_{1})\times...\times Z(G_{k}).$$
-Lem. 2: Let $k\in\mathbb{N}$. For $1\leq j\leq k$ let $G_{j}$ be a group and $N_{j}\trianglelefteq G_{j}$. Then \begin{align*}&i.)\;N_{1}\times...\times N_{k}\trianglelefteq G_{1}\times...\times G_{k};\\&ii.)\;(G_{1}\times...\times G_{k})/(N_{1}\times...\times N_{k})\cong (G_{1}/N_{1})\times...\times(G_{k}/N_{k}).\end{align*}
The Problem:
Let $k\in \mathbb{N}$. Let $G=G_{1}\times ...\times G_{k}$ where the $G_{j}$ are groups. For each $G_{j}$, let $N_{j}\trianglelefteq G_{j}$. Then for $i\geq 0$ we have $$Z_{i}(G_{1}\times...\times G_{k})=Z_{i}(G_{1})\times...\times Z_{i}(G_{k}).$$
My Attempt at Solution:
We proceed by induction on $i\geq 0$.
For the Base Case, $i=0$ is the trivial case. Note the $i=1$ case is given by Lem. 1 above.
So let $i\geq 0$. Suppose $Z_{i}(G_{1}\times...\times G_{k})=Z_{i}(G_{1})\times...\times Z_{i}(G_{k}).$ Now observe \begin{align*}Z_{i+1}(G_{1}\times...\times G_{k})/Z_{i}(G_{1}\times...\times G_{k}) &= Z((G_{1}\times...\times G_{k})/Z_{i}(G_{1}\times...\times G_{k}))\\&=Z((G_{1}\times...\times G_{k})/(Z_{i}(G_{1})\times...\times Z_{i}(G_{k})))\quad(\text{using induction hypothesis})\\&- - -\end{align*}
Comments:
... and this is where I'm stuck. My gut instinct was to use Lem. 2 to say $Z((G_{1}\times...\times G_{k})/(Z_{i}(G_{1})\times...\times Z_{i}(G_{k})))\cong Z(G_{1}/Z_{i}(G_{1})\times...\times G_{k}/Z_{i}(G_{k})),$ but I think this is wrong for two reasons: 1.) the problem asks to show equality, not isomorphism; and 2.) going through Dummit & Foote, we have not yet proven each $Z_{i}(G)\trianglelefteq G$ (this fact is a result of the very first exercise of Sec. 6.1 but the result I am trying to prove was presented mid-section by our professor). Also, no where in the proof I started above did I use each $N_{j}\trianglelefteq G_{j}$ which is concerning.
Any help, hints, or comments would be greatly appreciated!
The right and simplest approach is to do this elementswise. Let me do it for $G_1 \times G_2$ and proving $Z_2(G_1 \times G_2)=Z_2(G_1) \times Z_2(G_2)$ and I leave it to you to apply induction on the number of factors and the members of the upper central series.
For all $g_1 \in G_1, g_2 \in G_2$ we have $$(x_1,x_2) \in Z_2(G_1 \times G_2) \iff \\ [(x_1,x_2),(g_1,g_2)] \in Z(G_1 \times G_2) \iff \\ [(x_1,x_2),(g_1,g_2)] \in Z(G_1) \times Z(G_2) \iff \\ ([x_1,g_1]),[x_2,g_2]) \in Z(G_1) \times Z(G_2) \iff \\ [x_1,g_1] \in Z(G_1) \text { and } [x_2,g_2] \in Z(G_2) \iff \\ x_1 \in Z_2(G_1) \text { and } x_2 \in Z_2(G_2) \iff \\ (x_1,x_2) \in Z_2(G_1) \times Z_2(G_2).$$ Can you finish?