How can I prove the following Liouville theorem without using the mean value property?

If $u$ is harmonic on $\mathbb{R}^n$ and $\int_{\mathbb{R}^n}|D u|^2 dx \leq C$ for some $C > 0$, then $u$ is constant.

The proof that I know indeed uses the mean value property for harmonic functions.

From the comments: is it rigorous to do it like this: $-\Delta u = 0 \implies \int_{\mathbb R^n} |\nabla u|^2 = 0$ (integrating by parts, hence $u$ is constant? It seems to easy, probably I'm missing something.

How can I prove the following Liouville theorem without using the mean value property?

If $u$ is harmonic on $\mathbb{R}^n$ and $\int_{\mathbb{R}^n}|\nabla u|^2 dx \leq C$ for some $C > 0$, then $u$ is constant.

The proof that I know indeed uses the mean value property for harmonic functions.

From the comments: is it rigorous to do it like this: $-\Delta u = 0 \implies \int_{\mathbb R^n} |\nabla u|^2 = 0$ (integrating by parts, hence $u$ is constant? It seems to easy, probably I'm missing something.

How can I prove the following Liouville theorem without using the mean value property?

If $u$ is harmonic on $\mathbb{R}^n$ and $\int_{\mathbb{R}^n}|D u|^2 dx \leq C$ for some $C > 0$, then $u$ is constant.

The proof that I know indeed uses the mean value property for harmonic functions.

How can I prove the following Liouville theorem without using the mean value property?

If $u$ is harmonic on $\mathbb{R}^n$ and $\int_{\mathbb{R}^n}|D u|^2 dx \leq C$ for some $C > 0$, then $u$ is constant.

The proof that I know indeed uses the mean value property for harmonic functions.

From the comments: is it rigorous to do it like this: $-\Delta u = 0 \implies \int_{\mathbb R^n} |\nabla u|^2 = 0$ (integrating by parts, hence $u$ is constant? It seems to easy, probably I'm missing something.