Definition 1:
Let $G$ be a group. The lower central series of $G$ is the sequence $\big(C^n(G)\big)_{n\ge 1}$ of subgroups of $G$ defined inductively by: $C^1(G)=G$ and $$C^{n+1}(G)=[G,C^n(G)],$$ where $[G,C^n(G)]$ denotes the subgroup generated by the commutators $[g,g']$ with $g\in G$ and $g'\in C^n(G)$.
Definition 2:
A group $G$ is called nilpotent if there exists an integer $n$ such that $C^{n+1}(G)=\{e\}$. The least integer $n$ such that $C^{n+1}(G)=\{e\}$ is called the nilpotency class of a nilpotent group $G$.
Now, let $(G_i)_{1\leq i\leq p}$ be a finite sequence of nilpotent groups. I have to show that the group $$\prod_{i=1}^p G_i$$ is also nilpotent.
Attempt: (not much of an attempt)
Let $(n_i)_{1\leq i\leq p}$ be the sequence of the nilpotency classes of the family $(G_i)_{1\leq i\leq p}$: i.e. $n_i\in\mathbb{N}$ is the least integer such that
$$C^{n_i+1}(G_i)=\{e_i\}.$$
I have to find a $k\in\mathbb{N}$ such that $C^{k+1}(\prod_{i=1}^pG_i)=\{(e_i)_{1\leq i\leq p}\}$.
I have no idea where to start or what strategy to employ. I only know that somehow I have to use the sequence of nilpotency classes $(n_i)_{1\leq i\leq p}$. Any hints?
