Use a proof by induction on $m$:
$(a)$ It seems to me, then, that you have your base case $m = 2$: $P(2)$,
$\quad$ though it suffices, for a base case, to prove it's (trivially) true for $m = 1$.
$\quad$ But the process used in class for proving it's true for $m = 2$ will be helpful when making the
$\quad$ inductive step of the proof.
$(b)$ Your inductive hypothesis $P(k)$ would be to assume that this is true for $m = k$.
$\quad$ That is, assume the truth of: $$P(K):\quad
\det \begin{bmatrix}
A_1 &* &* &* &* &* \\
0& A_2 &* &* &* &* \\
.& 0& A_3 &* &* &* \\
.& 0& 0 &... &* &* \\
.& 0& 0& & ... &* \\
0& .& ...& 0&0 & A_k
\end{bmatrix}
= \prod_{i=1}^k \det A_i
$$
$(c)$ Then, take the inductive step: you'll need to use the inductive hypothesis to prove that $P(k+1)$ is true: for $m = k + 1$...
That is, assuming $P(k)$ is true, prove:
$$
\det \begin{bmatrix}
A_1 &* &* &* &* &| &* \\
0& A_2 &* &* &* &| &* \\
.& 0& A_3 &* &* &| &* \\
.& 0& 0 &... &* &| &* \\
.& 0& 0& ...& A_k &| &* \\
\hline& & & &\\
0& 0& ...& 0&0 &| & A_{k+1}
\end{bmatrix}
= \prod_{i=1}^{k+1} \det A_i
$$
Note that to do this, you can partition the matrix into two block matrices on the diagonal,
$(1)$ one of which is triangular (block) matrix with $k$ sub-blocks $A_i$ for $1 < i < k$ along it's diagonal, for which the determinant you know from the inductive hypothesis (having assumed its truth), and
$(2)$ the other with one block on the diagonal which we call $A_{k+1}$.
Here is where you can use the proof used in class for $m = 2$
That is: $$\prod_{i=1}^{k+1} \det A_i = \left(\prod_{i=1}^{k} \det A_i\right)\cdot \det(A_{k+1})
$$
Then you will have shown that the determinant of a partitioned triangular matrix is product of the determinants of the block matrices on the diagonal.
Your idea is good. If $n$ is even, you will need $\frac{n}{2}$ swap operations
$$ n \leftrightarrow 1, (n-1) \leftrightarrow 2, \ldots, \frac{n}{2}+1 \leftrightarrow \frac{n}{2}. $$
If $n$ is odd, you will need $\frac{n-1}{2}$ swap operations. Try to check that this works for small values of $n$.
Best Answer
One has: $$J:=\begin{pmatrix}0 & A\\B & 0\end{pmatrix}=\begin{pmatrix}A& 0\\0 & B\end{pmatrix}\times\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}.$$ You only have to compute: $$\varepsilon:=\det\left(\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}\right).$$ Indeed, using the first equality, one has: $$\det(J)=\varepsilon\det(A)\det(B).$$ If $n$ is odd, the result appears to be false, you will get: $$\det(J)=-\det(A)\det(B).$$