Let $(M,g)$ be a compact Riemannian manifold. The Hodge Laplacian is defined as follow \begin{align} \Delta:\Omega ^k(M)&\to \Omega^k(M)\\ \omega&\mapsto d\delta \omega+\delta d\omega \end{align} and every $\omega \in \ker \Delta$ is called a harmonic $p$-form.
Is the following differential operator coincide with Hodge Laplacian? \begin{align} \Delta^2:\Omega ^k(M)&\to \Omega^k(M)\\ \omega&\mapsto (d\delta)^2 \omega+(\delta d)^2\omega \end{align} If the answer is negative then Is it generate new invariant such as betti numbers?
Remark. It is easy to see that if $\omega \in \ker \Delta$ then $\omega \in \ker \Delta^2.$
The kernel is the same because $\Delta$ is self-adjoint. Namely, denote by
$$ (\omega, \eta) = \int_{M} \left< \omega, \eta \right> \, dVol_g $$
the inner product on forms. Then $(\Delta \omega, \eta) = (\omega, \Delta \eta)$ for all $\omega, \eta \in \Omega^{*}(M)$ and so if $\Delta^2 \omega = 0$ then
$$ 0 = (\Delta^2 \omega, \omega) = (\Delta \omega, \Delta \omega) = \int_M \left< \Delta \omega, \Delta \omega \right> \, dVol_g $$
and so $\Delta \omega = 0$.