Showing a certain form is exact I'm trying to solve the following: 
Let $f: S^{2n - 1} \rightarrow S^n$ be a smooth map, and let $\omega$ be an n-form on $S^n$ such that $\int_{S^n} \omega = 1$. Show that $f^*\omega$ is exact, and if $f^*\omega = d\alpha$, then $\int_{S^{2n-1}} \alpha \wedge d\alpha$ is independent of the choice of $\omega$ and $\alpha$.
I don't really know how to go about answering this. If I knew $f^*\omega$ was closed, then it would be exact since $S^{2n - 1}$ has no nth cohomology. For the second part, if n was odd, then we would have $d(\alpha \wedge \alpha) = -2 \alpha \wedge d\alpha$, so $\alpha \wedge d\alpha$ is exact and hence 0 independent of $\omega$ and $\alpha$.  
 A: Note that $\omega$ is an $n$-form on $S^n$, an $n$-dimensional manifold. Therefore $d\omega = 0$ as it is a $(n + 1)$-form on $S^n$. So $\omega$ is closed. Since pullbacks commute with exterior differentiation, we have that
$$d(f^\ast \omega) = f^\ast d\omega = 0,$$
so $f^\ast \omega$ is closed as well. Now apply what you said about $S^{2n-1}$ having trivial cohomology in degree $n$.

Now to show that
$$\int_{S^{2n-1}} \alpha \wedge d\alpha$$
is independent of the choice of $\omega$ and $\alpha$, let $\omega' \in \Omega^n(S^n)$ be another $n$-form on $S^n$ such that
$$\int_{S^n} \omega' = 1$$
and let $\alpha' \in \Omega^{n-1}(S^{2n-1})$ be such that $d\alpha' = f^\ast \omega'$. We will show that
$$\int_{S^{2n-1}} \alpha \wedge d\alpha = \int_{S^{2n-1}} \alpha' \wedge d\alpha',$$
which implies the integral is independent of the choices.
Since $H^n(S^n; \mathbb{R}) \cong \mathbb{R}$, by deRham's theorem there is some $\tau \in \Omega^{n-1}(S^n)$ such that
$$\omega' = \omega + d\tau.$$
Now
$$d(\alpha' - \alpha - f^\ast \tau) = f^\ast(\omega' - \omega - d\tau) = 0$$
and $H^n(S^{2n-1}; \mathbb{R}) \cong 0$, so again by deRham's theorem there exists some $\eta \in \Omega^{n-2}(S^{2n-1})$ such that
$$\alpha' = \alpha + f^\ast \tau + d\eta.$$
Hence we have that
\begin{align*}
\alpha' \wedge d\alpha' & = (\alpha + f^\ast \tau + d\eta) \wedge (d\alpha + f^\ast d\tau) \\
 & = \alpha \wedge d\alpha + \alpha \wedge d(f^\ast \tau) + f^\ast(\tau \wedge (\omega + d\tau)) + d(\eta \wedge (d\alpha + f^\ast d\tau)) \\
 & = \alpha \wedge d\alpha + \alpha \wedge d(f^\ast \tau) + d(\eta \wedge (d\alpha + f^\ast d\tau)),
\end{align*}
where in going to the third line we used the fact that $f^\ast(\tau \wedge (\omega + d\tau)) = 0$ since $\tau \wedge (\omega + d\tau) = 0$ as it is a $(n + 1)$ form on $S^n$. Now, since
\begin{align*}
\alpha \wedge d(f^\ast \tau) & = -d(\alpha \wedge f^\ast \tau) + d\alpha \wedge f^\ast \tau \\
 & = -d(\alpha \wedge f^\ast \tau) + f^\ast (\omega \wedge \tau) \\
 & = -d(\alpha \wedge f^\ast \tau),
\end{align*}
where once again $f^\ast (\omega \wedge \tau) = 0$ since $\omega \wedge \tau$ is an $(n+1)$-form on $S^n$, we get that
$$\alpha' \wedge d\alpha' = \alpha \wedge d\alpha + d(-\alpha \wedge f^\ast \tau + \eta \wedge (d\alpha + f^\ast d\tau)).$$
So by Stokes' theorem,
\begin{align*}
\int_{S^{2n-1}} \alpha' \wedge d\alpha' & = \int_{S^{2n-1}} (\alpha \wedge d\alpha + d(-\alpha \wedge f^\ast \tau + \eta \wedge (d\alpha + f^\ast d\tau))) \\
 & = \int_{S^{2n-1}} \alpha \wedge d\alpha,
\end{align*}
showing that the integral is independent of the choices.

Remark: The number
$$H(f) = \int_{S^{2n-1}} \alpha \wedge d\alpha$$
is called the Hopf invariant of the map $f$. Here we showed that it only depends on $f$. You can also show that it only depends on the homotopy class of $f$, which isn't too difficult.
