I am trying to show the following lemma:
Lemma: Let $n,m \geq 1$ and consider the switch map $s: S^m \wedge S^n \to S^n \wedge S^m$, given by $[(x,y)] \mapsto [(y,x)]$. Then we have $\text{deg}(s)=(-1)^{mn}$ if we view $s$ as a map on $S^{m+n}$ via the canonical isomorphism $S^m \wedge S^n \cong S^{m+n}$.
I have come up with the following: we have a homeomorphism $S^m \cong S^1 \wedge ... \wedge S^1$, the $m$-fold smash product of $S^1$'s. Write $x = (x_1,...,x_m)$ and $y = (y_1,...,y_n)$. Then $s$ is just a map \begin{align} S^1 \wedge ... \wedge S^1 \wedge S^1 \wedge ... \wedge S^1 &\to S^1 \wedge ... \wedge S^1 \wedge S^1 \wedge ... \wedge S^1\\ [ (x_1,...,x_m,y_1,...,y_n) ] &\mapsto [ (y_1,...,y_n,x_1,...,x_m) ] \end{align} Define a map $\tau \in \text{GL}(m+n,\mathbb{R})$, given by $(z_1,z_2,...,z_{m+n}) \mapsto (z_2,...,z_{m+n},z_1)$. We get an induced map $\overline{\tau}$ on the smash product of $S^1$'s and moreover we see that $s = \overline{\tau} \circ \ldots \circ \overline{\tau}$, the $(m+n)$-fold composition. Next we notice, by induction for example, that $\tau$ is represented by the $(m+n)$-matrix $$ A_\tau = \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & & \vdots \\ 0 & 0 & 0 & 1 & & \vdots \\ \vdots& & & & \ddots & 0 \\ 0 & & & & & 1 \\ 1 & 0 & \cdots & \cdots & \cdots & 0 \end{pmatrix} $$ and in particular it is easy to show that $A_\tau$ is orthogonal and that $\det(A_\tau) = (-1)^{m+n-1}$. Finally, using that orthogonal maps have degree the determinant of their representing matrix, we can write $$ \deg(s) = \deg(\overline\tau \circ \cdots \circ \overline\tau) = \deg(\overline\tau)^m = \det(A_\tau)^m = ((-1)^{m+n-1})^m = (-1)^{mn}\cdot (-1)^{m^2-m} = (-1)^{mn}, $$ since $m^2-m$ is even for all natural numbers $m \geq 1$.
My question: Is my proof flawed in any way? It feels weird using linear algebra, just because I don't see it as often in algebraic topology anymore.