$\Delta u = u$ implies $u$ infinitely differentiable, without Sobolev techniques

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Suppose $u:\mathbb{R}^n \to \mathbb{R}$ is a $C^2$ function which satisfies $\Delta u = u$. Can we prove that $u$ is $C^{\infty}$ without using the elliptic regularity theorem or any machinery of Sobolev spaces or weak derivatives?

Some ideas I´ve benn trying:

I've been trying to use mollifiers but i cant make it. The idea of using mollifiers was to bring regularity to the problem but this didnt help me.

Another idea was considering $f(x) = u(x)$ and work with the problem $\Delta u = f$. Solving this using Green and later prove that my original function $u$ differ from this on a $C^{\infty}$ function, ideally on a harmonic function. In the end this way has a lot of problems.

Also I had tried to find a shorcut on the ideas of the regularity theorem and sobolev theory. I know this theorem could answer the question, but using a lot of intermidiate spaces like $H^m$. I think that maybe asumming $u \in C^2$ there exists something like a shortcut.

Any ideas would be helpful