Smash and join products of spheres

How to prove rigorously that $\mathbb S^n \wedge \mathbb S^m =\mathbb S^{n+m}$ and $\mathbb S^n \ast \mathbb S^m = \mathbb S^{n+m+1}$? And what intuition should i have for compute $\wedge,\ast$ for difficult spaces?

I will prove that $\mathbb{S}^n \wedge \mathbb{S}^m$ is homeomorphic to $\mathbb{S}^{n+m}$

(Note that throughout this post $(\cdot)^*$ refers to the one-point compactification of a topological space)

Theorem: Let $M$ be a compact manifold of positive dimension and let $p \in M$. Then $M$ is homeomorphic to the one-point compactification of $M \setminus \{p\}$.

Note that $\mathbb{S}^n$ is a compact manifold. Let $p$ be any point in $\mathbb{S}^n$, then utilizing the above theorem we get $$\mathbb{S}^n \cong \left(\mathbb{S}^n \setminus \{p\}\right)^*$$ where $\left(\mathbb{S}^n \setminus \{p\}\right)^*$ denotes the one-point compactification of $\mathbb{S}^n \setminus \{p\}$. Note also that $$\mathbb{S}^n \setminus \{p\} \cong \mathbb{R}^n.$$This then allows us to conclude that since the one-point compactification of a topological space is unique up to homeomorphism, it follows that $$\left(\mathbb{S}^n \setminus \{p\}\right)^* \cong \left(\mathbb{R}^n\right)^*$$ and thus we can conclude that $$\mathbb{S}^n \cong \left(\mathbb{R}^n\right)^*. \ \ \ \ \ \ \ \ (1)$$

But how does this help us? In order to prove the desired result, we need to make use of one further theorem.

Theorem: If $X$ and $Y$ are compact Hausdorff spaces then $$X \wedge Y \cong \left(X \setminus \{x\} \times Y \setminus \{x\}\right)^*$$ where $x$ and $y$ are chosen base points of $X$ and $Y$ respectively.

Since $\mathbb{S}^n$ and $\mathbb{S}^m$ are both compact Hausdorff spaces the above theorem implies that $$\mathbb{S}^n \wedge \mathbb{S}^m \cong \left(\mathbb{S}^n \setminus \{p\} \times \mathbb{S}^m \setminus \{q\}\right)^*$$ and since $\mathbb{S}^n \setminus \{p\} \cong \mathbb{R}^n$ and $\mathbb{S}^m \setminus \{q\} \cong \mathbb{R}^m$ and the fact that for topological spaces $X, Y, A, B$ if $X \cong A$ and $Y \cong B$ then $X \times Y \cong A \times B$ it follows that

\begin{align*} \mathbb{S}^n \wedge \mathbb{S}^m &\cong \left(\mathbb{R}^n \times \mathbb{R}^m\right)^* \ \ \text{by the above} \\ &\cong \left(\mathbb{R}^{n+m}\right)^* \ \ \ \ \text{(since } \mathbb{R}^n \times \mathbb{R}^m \cong \mathbb{R}^{n+m} \text{)} \\ &\cong \ \mathbb{S}^{n+m} \ \ \ \ \ \ \text{by $(1)$ above} \end{align*}