[proofplan]
The proof is a direct translation between exactness and vanishing of the de Rham cohomology class. If $\alpha$ is exact, then its class in $H^k_{\mathrm{dR}}(M)$ is zero, so its pairing with every homology class is zero. Conversely, if every period of $\alpha$ over a smooth singular $k$-cycle vanishes, then the de Rham class $[\alpha]$ lies in the left kernel of the pairing; the assumed nondegeneracy forces $[\alpha]=0$, which is precisely exactness.
[/proofplan]
[step:Translate exactness into the zero de Rham cohomology class]
Because $\alpha \in \Omega^k(M)$ is closed, it defines a de Rham cohomology class
\begin{align*}
[\alpha] \in H^k_{\mathrm{dR}}(M).
\end{align*}
By definition of de Rham cohomology,
\begin{align*}
H^k_{\mathrm{dR}}(M)
=
\frac{\ker\left(d:\Omega^k(M)\to \Omega^{k+1}(M)\right)}
{\operatorname{im}\left(d:\Omega^{k-1}(M)\to \Omega^k(M)\right)}.
\end{align*}
Therefore $\alpha$ is exact if and only if $[\alpha]=0$ in $H^k_{\mathrm{dR}}(M)$. For $k=0$, this uses the standard convention that $\operatorname{im}(d:\Omega^{-1}(M)\to \Omega^0(M))=\{0\}$.
[/step]
[step:Show that an exact form has zero period over every cycle]
Assume that $\alpha$ is exact. By the previous step, $[\alpha]=0$ in $H^k_{\mathrm{dR}}(M)$.
Let $c$ be a smooth oriented singular $k$-cycle in $M$, and let $[c] \in H_k(M;\mathbb{R})$ denote its smooth singular homology class. Write
\begin{align*}
\langle \cdot,\cdot\rangle_{\mathrm{dR}}: H^k_{\mathrm{dR}}(M) \times H_k(M;\mathbb{R}) \to \mathbb{R}
\end{align*}
for the de Rham integration pairing. Then
\begin{align*}
\langle [\alpha],[c]\rangle_{\mathrm{dR}}
=
\langle 0,[c]\rangle_{\mathrm{dR}}
=
0,
\end{align*}
where the last equality follows from bilinearity of the pairing. Thus every period of $\alpha$ over a smooth oriented singular $k$-cycle vanishes.
[guided]
Assume that $\alpha$ is exact. Exactness means that its de Rham cohomology class is the zero class:
\begin{align*}
[\alpha]=0 \in H^k_{\mathrm{dR}}(M).
\end{align*}
Now let $c$ be any smooth oriented singular $k$-cycle in $M$. The cycle $c$ determines a homology class
\begin{align*}
[c] \in H_k(M;\mathbb{R}).
\end{align*}
Write
\begin{align*}
\langle \cdot,\cdot\rangle_{\mathrm{dR}}: H^k_{\mathrm{dR}}(M) \times H_k(M;\mathbb{R}) \to \mathbb{R}
\end{align*}
for the de Rham integration pairing. Applying this pairing to the zero class gives
\begin{align*}
\langle [\alpha],[c]\rangle_{\mathrm{dR}}
=
\langle 0,[c]\rangle_{\mathrm{dR}}.
\end{align*}
Because the pairing is bilinear, pairing the zero cohomology class with any homology class gives zero:
\begin{align*}
\langle 0,[c]\rangle_{\mathrm{dR}}=0.
\end{align*}
Therefore
\begin{align*}
\langle [\alpha],[c]\rangle_{\mathrm{dR}}=0.
\end{align*}
Since $c$ was arbitrary, all periods of $\alpha$ over smooth oriented singular $k$-cycles vanish.
[/guided]
[/step]
[step:Use vanishing of all periods to force the cohomology class to vanish]
Conversely, assume that
\begin{align*}
\langle [\alpha],[c]\rangle_{\mathrm{dR}}=0
\end{align*}
for every smooth oriented singular $k$-cycle $c$ in $M$.
Let $\gamma \in H_k(M;\mathbb{R})$ be arbitrary. By definition of smooth singular homology, there exists a smooth singular $k$-cycle $c$ in $M$ such that $\gamma=[c]$. The assumed period condition gives
\begin{align*}
\langle [\alpha],\gamma\rangle_{\mathrm{dR}}
=
\langle [\alpha],[c]\rangle_{\mathrm{dR}}
=
0.
\end{align*}
Thus $[\alpha]$ pairs to zero with every class $\gamma \in H_k(M;\mathbb{R})$. By the assumed nondegeneracy of the de Rham pairing in the first factor, it follows that
\begin{align*}
[\alpha]=0 \in H^k_{\mathrm{dR}}(M).
\end{align*}
[guided]
Now assume that every period of $\alpha$ over a smooth oriented singular $k$-cycle vanishes:
\begin{align*}
\langle [\alpha],[c]\rangle_{\mathrm{dR}}=0
\end{align*}
for every such cycle $c$.
To use the nondegeneracy hypothesis, we must show that the cohomology class $[\alpha]$ pairs to zero with every homology class. Let
\begin{align*}
\gamma \in H_k(M;\mathbb{R})
\end{align*}
be arbitrary. Since $H_k(M;\mathbb{R})$ is smooth singular homology, the class $\gamma$ is represented by some smooth singular $k$-cycle $c$ in $M$, so
\begin{align*}
\gamma=[c].
\end{align*}
By the definition of the de Rham integration pairing,
\begin{align*}
\langle [\alpha],\gamma\rangle_{\mathrm{dR}}
=
\langle [\alpha],[c]\rangle_{\mathrm{dR}}.
\end{align*}
The assumed period condition makes the last expression zero:
\begin{align*}
\langle [\alpha],\gamma\rangle_{\mathrm{dR}}=0.
\end{align*}
Since $\gamma$ was arbitrary, $[\alpha]$ lies in the left kernel of the pairing
\begin{align*}
H^k_{\mathrm{dR}}(M) \times H_k(M;\mathbb{R}) \to \mathbb{R}.
\end{align*}
The theorem assumes that this left kernel is zero. Therefore
\begin{align*}
[\alpha]=0 \in H^k_{\mathrm{dR}}(M).
\end{align*}
[/guided]
[/step]
[step:Conclude exactness from the zero cohomology class]
From the previous step, $[\alpha]=0$ in $H^k_{\mathrm{dR}}(M)$. By the definition of de Rham cohomology, this means precisely that $\alpha$ belongs to
\begin{align*}
\operatorname{im}\left(d:\Omega^{k-1}(M)\to \Omega^k(M)\right).
\end{align*}
Hence $\alpha$ is exact. Combining this with the forward implication proves the equivalence.
[/step]
