Solvability of Poisson's equation for $p$-forms directly from Hodge decomposition If $M$ is a closed manifold, it's well known that $\Delta\phi=\alpha$, where $\alpha$ is a $p$-form, has a solution if and only if $\alpha$ is orthogonal to the space of harmonic $p$-forms. This is proved on page 224 of Warner, but the proof depends heavily on his approach to Hodge theory. There are other methods that avoid his main technical lemma 6.6. So, can one prove this fact using only the "standard" facts of the theory?
For functions it is easy, for one has $$\alpha=\delta \psi+h.$$
But $(\alpha,h)=0$, so $\alpha=\delta\psi$. Then decompose $\psi$ to get $\alpha=\delta d\zeta$ for some function $\zeta$. But on functions, $\Delta=\delta d$. But for $p$-forms in general one has $$\alpha=d\xi+\delta \psi+h,$$
since $\Omega^{p-1}$ is nontrivial. Using the decompositions again, one has
$$\alpha=d\delta \rho+\delta d\sigma,$$
where $\rho,\sigma$ are $p$-forms. I don't see how $\rho$ and $\sigma$ are related, but it looks close. 
 A: If by "standard" facts of the theory, you mean the fact that every smooth $p$-form $\alpha$ can be decomposed as $\alpha=d\xi+\delta\psi+h$ with $h$ harmonic, here's a way to get from there to where you want to go.
The key is to observe that when we decompose $\alpha$ as above, we can require in addition that $\delta\xi=0$ and $d\psi=0$. To see why, use the decomposition to write $\xi = d\beta+\delta\gamma+h_1$, and note that $d\xi = d\delta\gamma$, so we can replace $\xi$ by $\delta\gamma$. A similar argument applies to $\psi$.
Now suppose $\alpha$ is a $p$-form that is orthogonal to the harmonic forms. As you noted, we can write $\alpha = d\delta\rho + \delta d\sigma$. In addition, we can assume $d\rho=0$ and $\delta\sigma=0$. Thus
$$
\Delta(\rho+\sigma) = d\delta\rho + \delta d\rho + d\delta\sigma + \delta d\sigma = d\delta\rho + \delta d\sigma = \alpha.
$$
It has to be noted, however, that Warner's "technical lemma 6.6" is exactly the heart of the matter. This is the estimate that's required to show that $\Delta$ has closed range, which in turn is the key ingredient in proving the decomposition theorem. There's no way to prove the Hodge theorem without some form of that estimate. 
