Consider the following differential $(n-1)$-form $\omega$ on $M$: for $p\in M$ and $v_1,\dots,v_{n-1}\in T_pM$, define
$$
\omega(v_1,\dots,v_{n-1}) = [ \psi(p), \nu(p), d\psi(v_1),\dots,d\psi(v_{n-1})]
$$
where the square brackets denote the standard volume form in $\mathbb R^{n+1}$ (in other words, the determinant).

A simple computation shows that
$$
d\omega = n(1+H\langle\psi,\nu\rangle)dA .
$$
Now the formula follows from the fact that $\int_M d\omega=0$ by Stokes.

This works for any immersed orientable hypersurface. For a non-orientable one, consider the oriented double cover. Note that in the embedded case one may assume that $M$ is a submanifold and $\psi$ is the inclusion, and then the formula for $\omega$ gets simpler: just remove all $\psi$ and $d\psi$.

Unfortunately I don't remember where I have read this proof, it was too long ago. (And I think that that text was only for $n=2$ anyway.)