The regular nilpotent orbit is dense in the nilpotent cone

Step 1: reduction to Jordan forms. The closure $\overline{\mathcal{O}}$ of the orbit $\mathcal{O}=\mathcal{O}(J_n(0))$ is invariant under conjugation: if $M_k\to M$ with $M_k\in\mathcal{O}$ then $PM_kP^{-1}\to PMP^{-1}$ and $PM_kP^{-1}\in\mathcal{O}$. Every $N\in\mathcal{N}_n$ is conjugate to a Jordan form $J_\lambda=\operatorname{diag}\bigl(J_{\lambda_1}(0),\dots,J_{\lambda_r}(0)\bigr)$ with $\lambda_1+\dots+\lambda_r=n$. So it suffices to show $J_\lambda\in\overline{\mathcal{O}}$ for every such $\lambda$.

Step 2: perturbing a Jordan form. $J_\lambda$ has ones in the superdiagonal positions $(i,i+1)$ except at the $r-1$ positions between consecutive blocks, where it has zeros. For $\varepsilon>0$ let $J_\lambda(\varepsilon)$ be obtained by putting $\varepsilon$ at those $r-1$ positions. Then $J_\lambda(\varepsilon)\to J_\lambda$ as $\varepsilon\to0$.

Step 3: $J_\lambda(\varepsilon)\in\mathcal{O}$. $M=J_\lambda(\varepsilon)$ is strictly upper triangular, so nilpotent, and all $n-1$ entries $m_{i,i+1}$ are nonzero. The $(1,n)$ entry of $M^{n-1}$ is $\prod_{i=1}^{n-1}m_{i,i+1}\ne0$ (only the path $1\to2\to\dots\to n$ contributes in a strictly upper triangular product), so $M^{n-1}\ne0$. A nilpotent matrix with $M^{n-1}\ne0$ has a single Jordan block of size $n$, so $M$ is conjugate to $J_n(0)$.

Conclusion. $J_\lambda$ is a limit of elements of $\mathcal{O}$, hence in $\overline{\mathcal{O}}$; by Step 1, $\mathcal{N}_n\subseteq\overline{\mathcal{O}}$. Since $\mathcal{O}\subseteq\mathcal{N}_n$ and $\mathcal{N}_n$ is closed, $\overline{\mathcal{O}}=\mathcal{N}_n$.

Remark. Thus the nilpotent cone is the closure of a single orbit (the "regular" one), and all other nilpotent orbits lie in the boundary. Lesson. Perturb a Jordan form by linking blocks to increase the size of the largest block.