[proofplan]
We use $\mathbb{F}_2$ coefficients, where every manifold is automatically orientable. Poincaré duality pairs the Betti numbers $\beta_i = \dim_{\mathbb{F}_2} H_i(M; \mathbb{F}_2)$ symmetrically: $\beta_i = \beta_{d-i}$ where $d = 2n+1$ is odd. The Euler characteristic is $\chi(M) = \sum_{i=0}^{d} (-1)^i \beta_i$, and the symmetry $\beta_i = \beta_{d-i}$ pairs each term $(-1)^i \beta_i$ with the term $(-1)^{d-i} \beta_{d-i} = -(-1)^i \beta_i$ (since $d$ is odd), causing complete cancellation.
[/proofplan]
[step:Establish Poincaré duality over $\mathbb{F}_2$]
Let $d = \dim M = 2n+1$. Since [every bundle is $\mathbb{F}_2$-orientable](/theorems/2281), the manifold $M$ is $\mathbb{F}_2$-oriented. Since $M$ is compact, $H_c^k(M; \mathbb{F}_2) = H^k(M; \mathbb{F}_2)$. By [Poincaré Duality](/theorems/2291) applied with $R = \mathbb{F}_2$, we obtain isomorphisms
\begin{align*}
D_M: H^k(M; \mathbb{F}_2) \xrightarrow{\;\sim\;} H_{d-k}(M; \mathbb{F}_2)
\end{align*}
for all $k$.
[guided]
Why do we work over $\mathbb{F}_2$ rather than $\mathbb{Z}$? Because Poincaré duality requires an orientation, and not every manifold is $\mathbb{Z}$-orientable (e.g., the real projective plane $\mathbb{RP}^2$). However, [every bundle is $\mathbb{F}_2$-orientable](/theorems/2281) — over $\mathbb{F}_2$, the local homology group $H_d(M, M \setminus \{x\}; \mathbb{F}_2) \cong \mathbb{F}_2$ has a unique generator, so there is no orientation choice to make. This lets us apply [Poincaré Duality](/theorems/2291) to any compact manifold without any orientability hypothesis.
Since $M$ is compact, compactly supported cohomology agrees with ordinary cohomology: $H_c^k(M; \mathbb{F}_2) = H^k(M; \mathbb{F}_2)$. The duality isomorphism then reads $H^k(M; \mathbb{F}_2) \cong H_{d-k}(M; \mathbb{F}_2)$.
[/guided]
[/step]
[step:Relate cohomological and homological Betti numbers over $\mathbb{F}_2$]
Since $\mathbb{F}_2$ is a field, the universal coefficients theorem gives a natural isomorphism
\begin{align*}
H^k(M; \mathbb{F}_2) \cong \operatorname{Hom}_{\mathbb{F}_2}(H_k(M; \mathbb{F}_2), \mathbb{F}_2).
\end{align*}
All homology groups $H_k(M; \mathbb{F}_2)$ are finite-dimensional $\mathbb{F}_2$-vector spaces (since $M$ is compact), so $\dim_{\mathbb{F}_2} H^k(M; \mathbb{F}_2) = \dim_{\mathbb{F}_2} H_k(M; \mathbb{F}_2)$. Define the $\mathbb{F}_2$-Betti numbers $\beta_i := \dim_{\mathbb{F}_2} H_i(M; \mathbb{F}_2)$. Combining with the Poincaré duality isomorphism from the previous step:
\begin{align*}
\beta_i = \dim_{\mathbb{F}_2} H_i(M; \mathbb{F}_2) = \dim_{\mathbb{F}_2} H^i(M; \mathbb{F}_2) = \dim_{\mathbb{F}_2} H_{d-i}(M; \mathbb{F}_2) = \beta_{d-i}.
\end{align*}
[/step]
[step:Pair terms in the alternating sum to show $\chi(M) = 0$]
The Euler characteristic satisfies
\begin{align*}
\chi(M) = \sum_{i=0}^{d} (-1)^i \beta_i
\end{align*}
(the Euler characteristic computed over any field equals the topological Euler characteristic, by the [Euler Characteristic Equals Homological Euler Characteristic](/theorems/2266)). Since $d = 2n+1$ is odd, pair the term at index $i$ with the term at index $d - i$:
\begin{align*}
(-1)^i \beta_i + (-1)^{d-i} \beta_{d-i} = (-1)^i \beta_i + (-1)^{d-i} \beta_i = (-1)^i \beta_i \bigl(1 + (-1)^{d-2i}\bigr).
\end{align*}
Since $d = 2n+1$ is odd, $d - 2i$ is odd for all $i$, so $(-1)^{d-2i} = -1$, and each pair sums to zero. The indices $\{0, 1, \ldots, d\}$ partition into pairs $\{i, d-i\}$ with no index left over (since $d$ is odd, $i \neq d-i$ for all $i$). Therefore
\begin{align*}
\chi(M) = \sum_{i=0}^{d} (-1)^i \beta_i = 0.
\end{align*}
[guided]
The key observation is that when $d$ is odd, the pairing $i \leftrightarrow d-i$ has no fixed point: if $i = d - i$, then $d = 2i$, which is impossible since $d$ is odd. So every index $i \in \{0, 1, \ldots, d\}$ is paired with a distinct index $d - i$.
For each such pair, the Poincaré duality symmetry $\beta_i = \beta_{d-i}$ combines with the sign pattern to give
\begin{align*}
(-1)^i \beta_i + (-1)^{d-i} \beta_{d-i} = (-1)^i \beta_i + (-1)^{d-i} \beta_i = \beta_i \bigl((-1)^i + (-1)^{d-i}\bigr).
\end{align*}
Now $(-1)^i + (-1)^{d-i} = (-1)^i(1 + (-1)^{d-2i})$. Since $d$ is odd and $2i$ is even, $d - 2i$ is odd, giving $(-1)^{d-2i} = -1$, so $1 + (-1)^{d-2i} = 0$. Each paired contribution vanishes, and since all terms are paired, $\chi(M) = 0$.
What fails in even dimensions? If $d = 2n$, the middle index $i = n$ satisfies $i = d - i$, so the middle Betti number $\beta_n$ contributes $(-1)^n \beta_n$ without a cancelling partner. This is why even-dimensional manifolds can have non-zero Euler characteristic (e.g., $\chi(S^{2n}) = 2$).
[/guided]
[/step]
