Let $U\subset \mathbb R^n$ be a bounded domain with Lipschitz boundary. The eigenvalue counting function
\begin{align*}
N(\Lambda)=|\{k\in\mathbb N:\lambda_k\le \Lambda\}|
\end{align*}
has leading-order growth
\begin{align*}
N(\Lambda)\sim \frac{\omega_n}{(2\pi)^n}\mathcal L^n(U)\Lambda^{n/2}
\end{align*}
as $\Lambda\to\infty$, where $\omega_n=\mathcal L^n(B(0,1))$.