Is there a way simplify this product?
$$ \sin\left({n} \frac{\pi}{2}\right) \sin\left({n} \frac{\pi}{3}\right) \sin\left({n} \frac{\pi}{4}\right) ...\sin\left({n} \frac{\pi}{n-1}\right) $$
And, is this the correct way to write it? $$ \prod_{m=2}^{n-1} \sin\left(n \frac{\pi}{m}\right) $$
I'm not a professional so I'd appreciate a simple explanation.
I don't think so. Using de Moivre's formula $e^{ix} = \cos x + i \sin x$ your product is related to all the finite series:
$$ n\pi \times \left( \pm \frac{1}{2} \pm \frac{1}{3} \dots \pm \frac{1}{n-1} \right) \mod 1$$
Your product is a weighted sum of $e^{[\dots]}$ for all series of this kind, behaving like the Boltzmann partition function. In fact the numbers in question wrap around the number like $[0,1]$ in a somewhat random fashion.
It's unlikely there is any simplification unless we try to estimate this series.
$$ \sum e^{n \pi i \cdot \left( \pm \frac{1}{2} \pm \frac{1}{3} \dots \pm \frac{1}{n-1} \right)}$$
That average is very likely to be close to $0$ since these numbers are equidistributed on the unit circle.
A histogram shows a little bit of variance, but basically the same idea:
It's very difficult to say how close to $0$ this result is. However there may be classical results for estimating random-ish sums of this type.