I'll answer this question from the theoretical side. The exponential behavior follows simply from the Schrödinger equation. Consider the one-electron Schrödinger equation:
$$
(-\frac{1}{2}\nabla^2 + V(\mathbf{r}))\psi(\mathbf{r}) = \epsilon\psi(\mathbf{r}), \epsilon < 0
$$
At spatial points that are very far away from the nucleus, $V(\mathbf{r})\approx 0$, so that the asymptotic solution is given by
$$
-\frac{1}{2}\nabla^2\psi(\mathbf{r}) = \epsilon\psi(\mathbf{r}), \epsilon < 0
$$
This differential equation has basic solutions of the form
$$
\psi(\mathbf{r}) = Ce^{-\sqrt{-2\epsilon}\mathbf{k}\cdot\mathbf{r}}
$$
for some unit vector $\mathbf{k}$. The real asymptotic behavior of $\psi(\mathbf{r})$ is thus a linear combination of these basic solutions. The linear combination may bring a polynomial prefactor to the exponential, but will never alter the exponent. Thus we have not only proved the exponential behavior, but also derived the correct exponent $\alpha = \sqrt{-2\epsilon}$. For a multi-electronic, non-interacting system, the overall decay rate is governed by the slowest decaying orbital, i.e. the HOMO.

Of course, the real wavefunction can only be described by a multi-electron Schrödinger equation. But we can work on the equivalent Kohn-Sham system and show that the Kohn-Sham wavefunction decays at a rate given by the Kohn-Sham HOMO energy. By Janak's theorem, the Kohn-Sham HOMO energy is just the negative of the ionization potential of the exact system. To see this, consider a huge ensemble of $N$ identical, non-interacting molecules. If we remove one electron from the ensemble and let the hole delocalize evenly between all the molecules, then as $N\to +\infty$, the electron removal has a negligible impact on the electron density of any molecule (and therefore the Kohn-Sham potential of each molecule). Therefore under the Kohn-Sham framework we see that removing such an electron costs an energy of $-\epsilon_{\mathrm{HOMO}}$ (it does not matter whether the HOMO refers to that of the ensemble or that of a molecule, since their orbital energies are equal), since the electron is taken from an energy level whose energy is $\epsilon_{\mathrm{HOMO}}$ and the Hamiltonian is not changed in this process. On the other hand, from the perspective of the real system it is clear that the energy cost is equal to the first ionization energy of one of the molecules, $I$. Therefore we have $\epsilon_{\mathrm{HOMO}} = -I$, which means that the Kohn-Sham wavefunction decays like (again up to a possible polynomial prefactor; the precise determination of this polynomial prefactor is a much more difficult question)
$$
\psi(\mathbf{r}) = Ce^{-\sqrt{2I}\mathbf{k}\cdot\mathbf{r}}
$$
Although the Kohn-Sham wavefunction is fictional, its density is equal to the true multielectronic density, and in order for the true density to have the same asymptotic behavior as the Kohn-Sham density, the true wavefunction must have the same asymptotic behavior as the Kohn-Sham wavefunction. Q.E.D.