[proofplan] We use the lower central series to locate the last nonzero term before the series vanishes. Minimality of the nilpotency index gives a nonzero subspace $\gamma_c(\mathfrak g)$ whose bracket with all of $\mathfrak g$ is zero. This forces every element of $\gamma_c(\mathfrak g)$ to lie in the center, so the center contains a nonzero subspace. [/proofplan] [step:Choose the last nonzero term of the lower central series] Since $\mathfrak g$ is nilpotent, the set \begin{align*} A := \{m \in \mathbb N : \gamma_{m+1}(\mathfrak g)=0\} \end{align*} is nonempty. By the well-ordering property of $\mathbb N$, let $c \in A$ be its least element. Thus \begin{align*} \gamma_{c+1}(\mathfrak g)=0. \end{align*} We claim that $\gamma_c(\mathfrak g)\neq 0$. If $c=1$, then $\gamma_c(\mathfrak g)=\gamma_1(\mathfrak g)=\mathfrak g\neq 0$ by hypothesis. If $c>1$ and $\gamma_c(\mathfrak g)=0$, then $c-1 \in A$, since \begin{align*} \gamma_{(c-1)+1}(\mathfrak g)=\gamma_c(\mathfrak g)=0, \end{align*} contradicting the minimality of $c$. Hence $\gamma_c(\mathfrak g)\neq 0$. [/step] [step:Show the last nonzero term is central] By the defining recursion for the lower central series, \begin{align*} [\mathfrak g,\gamma_c(\mathfrak g)] = \gamma_{c+1}(\mathfrak g) = 0. \end{align*} Let $x \in \gamma_c(\mathfrak g)$ and let $y \in \mathfrak g$. The equality above gives \begin{align*} [y,x]=0. \end{align*} Since the Lie bracket is alternating and bilinear, it is skew-symmetric, so \begin{align*} [x,y]=-[y,x]=0. \end{align*} Because this holds for every $y \in \mathfrak g$, the definition \begin{align*} Z(\mathfrak g):=\{z \in \mathfrak g : [z,y]=0 \text{ for every } y \in \mathfrak g\} \end{align*} implies $x \in Z(\mathfrak g)$. Therefore \begin{align*} \gamma_c(\mathfrak g)\subseteq Z(\mathfrak g). \end{align*} [/step] [step:Conclude that the center is nonzero] The subspace $\gamma_c(\mathfrak g)$ is nonzero and is contained in $Z(\mathfrak g)$. Therefore $Z(\mathfrak g)$ contains a nonzero element, and hence \begin{align*} Z(\mathfrak g)\neq 0. \end{align*} This proves the theorem. [/step]