Just want to provide details on the hint, and any corrections welcome.
Let $\omega$ be a smooth orientation form on $S^n$. Theorem 17.21 tells us that $[\omega]$ spans $H_{dR}^n \cong \Bbb{R}$. Assume $[\eta] \neq 0$. Then $[\eta] = c[\omega]$ for some scalar $c$. Thus, $[\eta] = [c\omega]$, which is to say $\eta = c\omega + \alpha$ for $\alpha$ exact. Then, since we've already shown exactness implies an integral of zero, we get
$$\int_{S^n}\eta = c\int_{S^n} \omega + \int_{S^n} \alpha = c\int_{S^n}\omega \neq 0.$$
Hence, $\int_{S^n}\eta = 0 \Rightarrow [\eta] = 0$, which means $\eta$ is exact.
The answer is already briefly described in the comment:
First, one can show that there is $a\in \mathbb R$ such that
$$ \int_M f^* \omega = a \int_N \omega$$
for all top form $\omega$ on $N$. This is already non-trivial and is proved here.
Now we see that $a =\mathrm{deg}(f)$. Let $q\in N$ be a regular value of $f$. Then for each $p\in f^{-1}(q)$, there is an open set $U_p $ of $M$ so that $f|_{U_p} \to f(U_p)$ is a diffeomorphism onto an open set $V_p = f(U_p)$. Then $f^{-1}(p)$ is discrete, and thus is finite since $M$ is compact. Write $f^{-1}(q) = \{p_1, \cdots, p_k\}$.
Then one can find a local chart $(V, \psi)$ centered at $q$ and disjoint open subsets $U_1, \cdots, U_k$ of $M$ so that $f^{-1}(V) = U_1\cup\cdots \cup U_k$ and $f|_{U_i} : U_i \to V$ is a diffeomorphism for each $i=1, \cdots, k$.
Let $\omega$ be a bump form (see e.g. here) on $N$, compactly supported in $V$ and $\int_N \omega \neq 0$. Then $f^*\omega$ is supported in $f^{-1}(V)$ (see here) and thus
$$ \int_M f^*\omega = \int_{f^{-1}\ (V)} f^*\omega = \sum_i \int_{U_i} f^*\omega.$$
By the change of variable formula, since each $f|_{U_i}$ is a diffeomorphism,
$$ \int_{U_i} f^*\omega = \pm\int_V \omega $$
where $\pm$ depends if $df_{p_i} :T_{p_i}M \to T_qN$ is orientation preserving or reversing. Thus
$$ \int_M f^*\omega = \mathrm{deg}(f) \int_V\omega = \mathrm{deg}(f) \int_N \omega.$$
Since this is true for this $\omega$ and $\int_N \omega \neq 0$, $a=\mathrm{deg}(f)$ and thus the same holds for all top forms.
Best Answer
Let $(U, \varphi)$ be an oriented local chart of $M$. On $\varphi (U) \subset \mathbb R^k$ you have a $k$-form $\omega_0 = dx^1 \wedge \cdots \wedge dx^k$ and a bump function $b : \varphi(U) \to \mathbb R$ (i.e. a smooth function with compact support in $\varphi(U)$), then $$\varphi^*(b \omega_0)$$ is a smooth $k$-form on $M$ (by extending to zero outside of $U$). By definition of the integration,
$$ \int_M \omega = \int_{\varphi (U)} b\omega_0 = \int_{\varphi (U)} b(x) dx^1\cdots dx^k.$$
One can choose this to be non-zero (e.g. choose $b\ge 0$ and $b>0$ on an open subset)