In relation to this question, consider the quantity
$$\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n$$
when $n \in \mathbb{N}$ and $n \geq 2$.
Under these conditions, it should be a real number.
Is it possible to exploit the presence of $n$ both in the denominator and in the power, to further simplify the expression?
My attempt:
$$\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n = \left( e^{i \frac{\pi}{2n}} + e^{-i \frac{\pi}{2n}} \right)^n = \left( i^{\frac{1}{n}} + \frac{1}{i^{\frac{1}{n}}} \right)^n = \left( \frac{i^{\frac{2}{n}} + 1}{i^{\frac{1}{n}}} \right)^n$$
But I am stuck here and I am no more sure that this is a real quantity.
Edit: trying to evaluate $2 \cos \left( \frac{\pi}{2n} \right)$ for some values of $n$, I obtain
$$n = 2; \ 2 \frac{\sqrt{2}}{2} = \sqrt{2}\\ n = 3; \ 2 \frac{\sqrt{3}}{2} = \sqrt{3}\\ n = 4; \ 2 \frac{\sqrt{2 + \sqrt{2}}}{2} = \sqrt{2 + \sqrt{2}}\\ n = 5; \ 2 \frac{\sqrt{5 + \sqrt{5}}}{2 \sqrt{2}} = \frac{\sqrt{5 + \sqrt{5}}}{\sqrt{2}}\\ n = 6; \ 2 \frac{\sqrt{2 + \sqrt{3}}}{2} = \sqrt{2 + \sqrt{3}}$$
I'm not sure about it being a regular succession. The $n = 5$ term is confusing.
As regards the binomial theorem applied to $\left( e^{i \frac{\pi}{2n}} + e^{-i \frac{\pi}{2n}} \right)^n$, I don't think to be able to manage those terms in the attempt to identify something useful.
Don't see a reason to write in terms of imaginary numbers.
Distribute the power of $n$: $$\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n$$ $$= 2^n \left[\cos \left( \frac{\pi}{2n} \right) \right]^n$$
Using the half-angle identity Gerry mentioned: $$\cos(\frac{x}{2})=\sqrt{\frac{1+\cos x}{2}}$$ Where $x=\pi/n$: $$\cos(\frac{\pi}{2n})=\sqrt{\frac{1+\cos {\frac{\pi}{n}}}{2}}$$ Subbing back in: $$= 2^n \left[\frac{1+\cos {\frac{\pi}{n}}}{2} \right]^{n/2}$$ Since $1+\cos {\theta}$ is always positive, this expression will result in a real number.