Let $P$ and $Q$ be the products of the sines of the odd and even degree angles up to $89^\circ;$ pair $\sin k^\circ$ with $\sin(90-k)^\circ = \cos k^\circ$

Then $P^2Q^2 = \prod \sin k^\circ \cos k^\circ;$ doubling every factor turns the product into $Q^2$ again, leaving only a power of $2$

Let $P = \sin 1^\circ \sin 3^\circ \cdots \sin 89^\circ$ and $Q = \sin 2^\circ \sin 4^\circ \cdots \sin 88^\circ,$ so the desired product is $\frac{1}{P^2}.$ Then $PQ = \prod_{k=1}^{89} \sin k^\circ,$ and multiplying this by itself in reverse order, using $\sin(90 - k)^\circ = \cos k^\circ,$ gives $$P^2Q^2 = \prod_{k=1}^{89} \sin k^\circ \cos k^\circ.$$

Multiply by $2^{89}$ and use $2\sin k^\circ \cos k^\circ = \sin 2k^\circ:$ $$\begin{aligned} 2^{89} P^2 Q^2 &= \prod_{k=1}^{89} \sin 2k^\circ \\ &= \left(\prod_{k=1}^{44} \sin 2k^\circ\right) \\ &\quad {}\cdot \left(\prod_{k=46}^{89} \sin 2k^\circ\right) \\ &= Q \cdot Q, \end{aligned}$$ since $\sin 90^\circ = 1$ and $\sin(180 - x)^\circ = \sin x^\circ$ turns the second half into $Q$ as well.

Because $Q \ne 0,$ it follows that $P^2 = 2^{-89},$ so $\prod_{k=1}^{45} \csc^2(2k-1)^\circ = 2^{89}.$ Since $89$ is prime, the only representation $m^n$ with $m, n \gt 1$ is $m = 2,$ $n = 89,$ and $m + n = 91.$