The unitriangular group $UT_n(\Bbb Z)$ is the group of all $n \times n$ invertible triangular matrices with the identity on each entry of the main diagonal, and integer entries everywhere else in the triangle. Show that this group is nilpotent, and that its nilpotence class is $n$.
Definition (upper central series): For any group $G$ define the following subgroups inductively: $$Z_0(G) = 1, \qquad Z_1(G) = Z(G)$$ and $Z_{i+1}(G)$ is the subgroup of $G$ containing $Z_i(G)$ such that $$Z_{i+1}(G)/Z_i(G) = Z(G/Z_i(G)).$$ The chain of subgroups $$Z_0(G) \leq Z_1(G) \leq Z_2(G) \leq \cdots$$ is called the upper central series of $G$.
Definition (nilpotent): A group $G$ is called nilpotent if $Z_c(G) = G$ for some $c \in \Bbb Z$. The smallest such $c$ is called the nilpotence class of $G$.
To show that it is nilpotent, I think it is suffient to show that it is a $p-$group; i.e. $|UT_n(\Bbb Z)| = p^{\alpha}$ where $p$ is a prime number and $\alpha$ is a positive integer. I feel like there must be some sort of algorithm to calculate how many possible matrices we can get, similar to the formula for finding the order of $GL_n(\Bbb F)$, the general linear group. I tried Googling, but I can't find a formula for the unitriangular group $UT_n(\Bbb Z)$.
To show the nilpotence class is $n$, I have to prove that $Z_n(G) = G$, and that $n$ is the smallest such integer. So by the given definition above, I know that $Z_n(G)/Z_{n-1}(G) = Z(G/Z_n(G))$. How can I manipulate this to arrive at $Z_n(G) = G$, and also show that $n$ is the smallest such integer?