Let \begin{equation*} 1=Z_0(G)\lhd Z_1(G)\lhd ...\lhd Z_{n-1}(G)\lhd Z_n(G)=G \end{equation*} be the upper central series of a connected nilpotent Lie group $G$, i.e. $Z_i(G)=\{g\in G:[g,G]\subset Z_{i-1}(G)\}$, $i=1,...,n$.
Question: Are the $Z_i(G)$ Lie groups? If so, do we have $\text{Lie}(Z_i(G))=\mathfrak{z}_i(\mathfrak{g})$, where \begin{equation*} 0=\mathfrak{z}_0(\mathfrak{g})\lhd\mathfrak{z}_1(\mathfrak{g})\lhd...\lhd\mathfrak{z}_{n-1}(\mathfrak{g})\lhd\mathfrak{z}_n(\mathfrak{g})=\mathfrak{g} \end{equation*} is the upper central series of $\mathfrak{g}=\text{Lie}(G)$? Any references would be helpful!
My thoughts: I know that the analogous statement is true for the lower central series. Also, I was able to prove that if \begin{equation*} 1=N_0\lhd N_1\lhd ...\lhd N_{n-1}\lhd N_n=G \end{equation*} is a central series for $G$, where the $N_i$ are Lie groups, then the corresponding Lie algebras form a central series for $\text{Lie}(G)$. When trying to prove the statement for the upper central series, I tried to use induction but I got stuck. My Ansatz would work out if $\text{Lie}(\varphi^{-1}(Z(G)))=\text{Lie}(\varphi)^{-1}(\mathfrak{z}(\mathfrak{g}))$, where $\varphi$ is a surjective Lie group homomorphism. However, I don't know how to prove this... (Edit: I'm also aware of the closed subgroup theorem, so to show that the $Z_i(G)$ are Lie groups, it would be enough to show that they are closed.)