The singular homology and cohomology of manifolds vanishes in high dimensions
In Algebraic Topology by Allen Hatcher:
Theorem 3.26. Let $M$ be a closed, connected $n$-manifold. Then:
(a) If $M$ is $R$-orientable, the map $H_n(M;R) \to H_n(M | x; R) \approx R$ is an isomorphism for all $x \in M$.
(b) If $M$ is not $R$-orientable, the map $H_n(M; R) \to H_n(M | x; R) \approx R$ is injective with image $\{ r \in R | 2r = 0 \}$ for all $x \in M$.
(c) $H_i(M; R) = 0$ for $i > n$.
Here $H_k(X | A; G) := H_k(X, X \setminus A; G)$. The proof is somewhat technical and I suggest you read the book to get a full idea of how it works.
The basic idea is that you want to show $H_i(M | A) = 0$ for $i > n$ when $A$ is a compact subset, and then take $A = M$. First you show that if it holds for $A$, $B$ and $A \cap B$ then it holds for $A \cup B$. Now you use that $M$ is a manifold and the previous property to reduce to the case $M = \mathbb{R}^n$. You reduce then to the case of a simplicial complex $A = K$, and then it follows by induction on the number of simplices.
For cohomology, now use the universal coefficient theorem, and the explicit description of $H_n(M;\mathbb{Z})$ given by the theorem depending on whether $M$ is $\mathbb{Z}$-orientable or not. For $k > n$, you get an exact sequence: $$0 \to \operatorname{Ext}^1_\mathbb{Z}(H_{k-1}(M; \mathbb{Z}), R) \to H^k(M; R) \to \hom_\mathbb{Z}(H_k(M; \mathbb{Z}), R) \to 0.$$
Since $H_{k-1}(M; \mathbb{Z})$ is either $0$ (when $k > n+1$) or a subgroup of $H_n(M; \mathbb{Z})$ (when $k = n+1$), and is therefore free abelian (because $H_n(M; \mathbb{Z})$ is either $\mathbb{Z}$ or zero depending on the orientability of $M$), it is always projective as a $\mathbb{Z}$-module, so the $\operatorname{Ext}$ vanishes. Meanwhile $H_k(M; \mathbb{Z})$ is zero ($k > n$). In the end the left and right terms are zero so the middle term $H^k(M; R)$ is zero too.
For noncompact manifolds, there is the following proposition:
Proposition 3.29. If $M$ is a connected noncompact $n$-manifold, then $H_i(M; R) = 0$ for $i \ge n$.
The proof is basically a corollary of the compact case (I'll just give the general idea here). Let $[z] \in H_i(M; R)$ is represented by a cycle $z$, then $z$ has compact image in $M$ (because $\Delta^i$ is compact). So you can find a relatively compact open set $U$ containing the image of $z$; let $V = M \setminus \overline{U}$.
For $i > n$ you can simply apply the long exact sequence of the triple $(M, U \cup V, V)$ to get that $[z] = 0$. For $i = n$ it's more complicated and you have to reuse the proof of the previous theorem, and use that since $M$ is noncompact there are points not in the image of $z$. See the book for more details.
Now for a "dumb" proof: use Poincaré duality, then if $i > n$, you find $H^i(M) \cong H_{n-i}(M) = 0$ because $n-i < 0$. I say "dumb" because the proof of the Poincaré duality theorem depends on the fact that $H_*$ vanishes in dimensions higher that $n$, so the proof is circular... It could still be a helpful mnemonic, and it gives a hint that for extraordinary cohomology theories, the result isn't necessarily true (indeed $K^{\mathbb{C}}_{2n}(*) \neq 0$ for example).