Knowing that $\prod_{n=1}^{\infty}\text{cos}\frac{x}{2^n} = \frac{\text{sin }x}{x}$, I am looking into the properties of $f(x)=\prod_{n=1}^{\infty}\text{cos}\frac{x}{n}$ to see if it can be simplified as well; Taking the natural log and comparing the summands to $-\frac{x^2}{n^2}$ will show the infinite product converges for all $x\in\mathbb{R}$ by the Direct Comparison Test.
I think the derivatives are very fascinating. Using logarithmic differentiation will show that the first three derivatives are:
$$f'(x)=-f\left(x\right)\sum_{n=1}^{\infty}\frac{\tan\left(\frac{x}{n}\right)}{n}$$ $$f''(x)=f\left(x\right)\left(\left(\sum_{n=1}^{\infty}\frac{\tan\left(\frac{x}{n}\right)}{n}\right)^{2}-\sum_{n=1}^{\infty}\frac{\sec^2\left(\frac{x}{n}\right)}{n^2}\right)$$ $$f'''(x)=f\left(x\right)\left(3\left(\sum_{n=1}^{\infty}\frac{\tan\left(\frac{x}{n}\right)}{n}\right)\left(\sum_{n=1}^{\infty}\frac{\sec^2\left(\frac{x}{n}\right)}{n^2}\right)-\left(\sum_{n=1}^{\infty}\frac{\tan\left(\frac{x}{n}\right)}{n}\right)^{3}-2\sum_{n=1}^{\infty}\frac{\sec^{2}\left(\frac{x}{n}\right)\tan\left(\frac{x}{n}\right)}{n^{3}}\right)$$ $f(0)=1$, and $f'(0)$ and $f'''(0)$ will both result in $0$ (it is likely that $f^{(2n+1)}(0)=0$ for integers $n$). However, $f''(0)=-\zeta(2)=-\frac{\pi^2}{6}$, and although I can't type the fourth derivative, $f^{(4)}(0)=3\zeta^2(2)-2\zeta(4)=\frac{11\pi^4}{180}$. My hypothesis is that since the derivatives of the secant and tangent result in each other, the Maclaurin coefficients will purely consist of the zeta function at integers.
To make this a question, I'll ask two; could there be a pattern to $f^{(n)}(0)$? Might they contain the odd arguments of the zeta function or not? I think proving that $f^{(2n+1)}(0)$ = $0$ would help answer the second one, but I don't know how to do that.
I suppose that we could use $$\log (\cos (t))=\sum_{k=1}^\infty (-1)^k \,\frac{ 2^{2 k-1} \left(4^k-1\right) B_{2 k}}{k\, (2 k)!}\,t^{2k}$$ where $B_n$ is the Bernoulli number.
Let $t=\frac x n$, switch summations to obtain $$\color{blue}{\log(f(x))=-\sum_{k=1}^\infty \,4^{k-1} \left(4^k-1\right)\, \frac{ \left|B_{2 k}\right|\, \zeta (2 k)}{ \,k^2\, \Gamma (2 k)}\,x^{2k}}$$