<s>[INST] Simplify $\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n$ 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.
 [/INST] 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.
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