De Rham cohomology of compact manifold minus one point

The trick is to look at the actual arrows involved. Also for convenience sake, I will assume $N$ is actually $M$ delete an open ball, since they're homotopy equivalent.

The $0$, $1$ exact sequence

Consider the exact sequence $$ 0 \to H^0(M)\to H^0(N)\oplus \Bbb{R}\to \Bbb{R}\to H^1(M)\to H^1(N)$$ The trick is to observe that $H^0(M)=\Bbb{R}$, since $M$ is connected. Thus we have $$ 0 \to \Bbb{R} \to H^0(N)\oplus \Bbb{R}\to \Bbb{R},$$ and since $N\ne \varnothing$, we see that the dimension of $H^0(N) \oplus \Bbb{R}$ is greater than or equal to 2, but it's also at most 2, since it fits into this exact sequence. Thus $H^0(N)=\Bbb{R}$ and the map $H^0(N)\oplus \Bbb{R}\to \Bbb{R}$ is surjective.

We could also show this geometrically ($N$ must be connected, so $H^0(N)=\Bbb{R}$ from that consideration alone, and surjectivity of the map is clear geometrically).

Regardless, surjectivity of this map allows us to split the exact sequence as $$ 0\to H^0(M)\to H^0(N)\oplus \Bbb{R} \to \Bbb{R} \to 0$$ and $$ 0\to H^1(M)\to H^1(N).$$ If $n> 2$, then this fits in as $0\to H^1(M)\to H^1(N)\to 0$, giving $H^1(M)\simeq H^1(N)$. Otherwise this is part of the other exact sequence we don't understand, so let's look at that one next.

The $n$, $n-1$ exact sequence

We have $$ 0 \to H^{n-1}(M) \to H^{n-1}(N) \to \Bbb{R} \to H^n(M)\to H^n(N) \to 0$$

Consider the map $H^{n-1}(N)\to \Bbb{R}$. This is induced by restricting a differential form $\omega$ representing some cohomology class $[\omega]$, defined on $N$ to the boundary sphere of $N$.

Then observe that $$\int_{\partial N} \omega = \int_N d\omega = \int_N 0 = 0,$$ by Stokes' theorem and that $\omega$ is closed.

Therefore the image of $[\omega]$ in $\Bbb{R} = H^{n-1}(\partial N)$ is $0$. Thus the map $H^{n-1}(N)\to \Bbb{R}$ is $0$.

Therefore we can split the exact sequence again into the pieces $$0\to H^{n-1}(M)\to H^{n-1}(N)\to 0$$ and $$0\to \Bbb{R}\to H^n(M)\to H^n(N)\to 0,$$ and using that $H^n(M)=\Bbb{R}$, since $M$ is connected, compact, and orientable, we see that $H^n(N)=0$, as desired.

Alternatively without using that $H^n(M)=\Bbb{R}$, we already have $b_n(N) = b_n(M)-1$.

Let $U=N$, $V$ the domain of a chart which contains the point removed.

We obtain from Mayer Vietoris the following sequence is exact:

$H^{i-1}(U\cap V)\rightarrow H^i(X)\rightarrow H^i(U)\oplus H^i(V)\rightarrow H^i(U\cap V)$

Remark that $U\cap V$ is a deformation retract of $S^{n-1}$, thus $H^i(U\cap V)=0$ if $i\neq 0, n-1$ and $H^i(U)=0$ if $i\neq 0$ since $U$ is diffeomorphic to $\mathbb{R}^n$, this implies that if $i<n-2$ we have

$0=H^{i-1}(U\cap V)\rightarrow H^i(X)\rightarrow H^i(U)\oplus (H^i(V)=0)\rightarrow H^i(U\cap V)=0$

implies that $H^i(X)=H^i(N)=H^i(N)$.

For $i=n-1$,

$0=H^{n-2}(U\cap V)\rightarrow H^{n-1}(X)\rightarrow H^{n-1}(U)\oplus H^{n-1}(V)\rightarrow (H^{n-1}(U\cap V)=\mathbb{Z})$

$\rightarrow( H^n(X)=\mathbb{Z})\rightarrow H^n(U)\oplus (H^n(V)=0)\oplus H^n(U\cap V)=0$

implies that $H^{n-1}(X)=H^{n-1}(V)=H^{n-1}(N)$.

Here I use the fact that the top cohomology group of a manifold without a point vanishes which is shown here:

Top homology of an oriented, compact, connected smooth manifold with boundary