I would like to know if there is a closed form formula for the homotopy type of $\widehat{(\mathbb{C^{\ast}})^n}$? For example, it is not difficult to see that $\widehat{\mathbb{C^{\ast}}}$ has the homotopy type of $S^1\vee S^2$.


My guess is that the formula, for general $n$, should look something like this
$$ \bigvee\_{i=0}^n \left(\bigvee\_{\binom{n}{i}} S^{n+i}\right). $$
The hunch is based on the fact that the compactly supported cohomology of the $n$-torus has the homology of the above wedge of spheres. However, I am not sure how I can prove this. 


Note: I did post the same question on Math StackExchange; here is the [link](https://math.stackexchange.com/questions/2732057/one-point-compactification-of-a-complex-torus). 



One point compactification of $(\mathbb{C}^{\ast})^n$
If $\widehat X$ is the 1-point compactification of $X$, then there is a homeomorphism (for, say, locally compact Hausdorff spaces)
$$
\widehat{X \times Y} \cong \widehat X \wedge \widehat Y
$$
with the smash product. Moreover, the smash product preserves homotopy equivalences for well-pointed spaces, which $\widehat{\Bbb C^\*}$ and $S^1 \vee S^2$ both are.


Therefore, because the smash product distributes over the wedge product, there is a homotopy equivalence
$$
\begin{align\*}
\widehat{(\Bbb C^\*)^n} &\simeq (S^1 \vee S^2)^{\wedge n}\\
&\cong \bigvee\_{i\_k \in \{1,2\}} S^{i\_1} \wedge S^{i\_2} \wedge \dots \wedge S^{i\_n}\\
&\cong \bigvee\_{i=0}^{n} \bigvee\_{\binom{n}{i}} S^{n+i}
\end{align\*}
$$
 as desired.


