Closed Form of the Real Portion of $f(n) = \prod_{m=2}^{n-1} e^{\pi i n/m}$

Question

I am wondering if it is possible to express an equation in closed form. I currently have:

$$f(n) = \prod_{m=2}^{n-1} e^{\pi i n/m}$$

Where $i$ is the $\sqrt{-1}$, which I know it commonly represents but due to the finite product I figured I would add this note for clarity.

Context: I am trying to work towards a solution for Simplify Product of sines, yet so far I have only arrived here.

Some options I have been considering to try and find a closed form include cases such as I only actually care when $n$ is an odd number so discarding even cases is fine if that simplifies it. I am also fine with having the series start a $m=1$ instead of $m=2$ if that somehow simplifies the answer. The other potential saving grace is that I may only care about the real portion of the answer and not the imaginary portion if that simplifies things.

Sorry I am a computer scientist not a mathematician so please let me know if I should adjust the title or the wording of the question for clarity. Any leads or help of any form would be appreciated as I am currently lost.

Answer 1

$$ \begin{align} \prod_{m=2}^{n-1}e^{\frac{\pi in}m} &=e^{\pi in(H_{n-1}-1)}\\ &=(-1)^ne^{\pi inH_{n-1}}\\[9pt] &=(-1)^{n-1}e^{\pi inH_n}\tag1 \end{align} $$ where $H_n$ is the $n^\text{th}$ Harmonic Number. $$ H_n\sim\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\frac1{252n^6}+\frac1{240n^8}-\frac1{132n^{10}}\tag2 $$ and $\gamma$ is the Euler-Mascheroni constant.

The real portion of $(1)$ is $$ \operatorname{Re}\left(\prod_{m=2}^{n-1}e^{\frac{\pi in}m}\right)=(-1)^{n-1}\cos\left(\pi nH_n\right)\tag3 $$

Answer 2

$$\begin{align} f(n)=&\prod_{m=2}^{n-1}\exp\frac{i\pi n}{m}\\ &=\exp\left[\sum_{m=2}^{n-1}\frac{i\pi n}{m}\right]\\ &=\exp\left[i\pi n\sum_{m=2}^{n-1}\frac{1}{m}\right]\\ \end{align}$$ Recalling the definition of the harmonic numbers: $$H_n=\sum_{m=1}^n\frac1m$$ We have that $$f(n)=\exp\left[i\pi n(H_{n-1}-1)\right]$$ Then using $e^{i\theta}=\cos\theta+i\sin\theta$, $$f(n)=\cos\left[\pi n(H_{n-1}-1)\right]+i\sin\left[\pi n(H_{n-1}-1)\right]$$ So $$\text{Re}f(n)=\cos\left[\pi n(H_{n-1}-1)\right]$$