Pullback of $S^n$ volume form

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I am working on problems from Nakahara's book Geometry, Topology and Physics and I am struggling with exercise 9.2

Let $\Omega_n$ be the volume form of $S^n$ normalized to 1, $f:S^{2n-1}\to S^n$ be a smooth map and denote by $f^*\Omega_n$ the pullback of $\Omega_n$ to $S^{2n-1}$.
(a) Show that $f^*\Omega_n$ is closed and written as $d\omega_{n-1}$ where $\omega_{n-1}$ is an $(n-1)$-form on $S^{2n-1}$.

I showed that $\Omega_n$ is closed, since $df^*\Omega_n=f^*d\Omega_n$ and $\Omega_n$ is a top form in $S^n$. I cannot show that $f^*\Omega_n$ is exact though. Indeed, I would tend to think that it is not. This is my argument.\

By Hodge's decomposition, I can write \begin{equation} \Omega_n=d\alpha_{n-1}+\gamma_n \end{equation} where $\gamma_n$ is harmonic ($\Delta \gamma_n=0$). The pullback of $\Omega_n$ is thus \begin{equation} f^*\Omega_n=f^*d\alpha_{n-1}+f^*\gamma_n=df^*\alpha_{n-1}+f^*\gamma_n \end{equation} Now, by the uniqueness of the decomposition, if $f^*\Omega_n$ were to be exact, I should have $f^*\alpha_{n-1}=\omega_{n-1}$ and $f^*\gamma_n=0$.
However, $\gamma_n$ is different from $0$ as $\Omega_n$ is not exact, so $f^*\gamma_n\ne 0$.

Where am I wrong?