\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We notice \[ (z - \exp (i\theta )) (z - \exp (-i\theta )) = z^2 - (\exp (i\theta ) + \exp (-i\theta ))z + 1 = z^2 - 2z \cos \theta + 1. \]
The \(2n\)-th roots of \(-1\) are \(z_r\), where \(r = 0, 1, \ldots , 2n - 1\), \[ z_r = \exp \left (i\left (\frac {\pi }{2n} + \frac {2r\pi }{2n}\right )\right ) = \exp \left (i\pi \cdot \frac {1 + 2r}{2n}\right ), \] and hence \begin {align*} z^{2n} + 1 & = \prod _{r = 0}^{2n - 1} (z - z_r) \\ & = \left [\prod _{r = 0}^{n - 1} \left (z - \exp \left (i\pi \cdot \frac {1 + 2r}{2n}\right )\right )\right ] \cdot \left [\prod _{r = n}^{2n - 1} \left (z - \exp \left (i\pi \cdot \frac {1 + 2r}{2n}\right )\right )\right ] \\ & = \left [\prod _{r = 0}^{n - 1} \left (z - \exp \left (i\pi \cdot \frac {1 + 2r}{2n}\right )\right )\right ] \cdot \left [\prod _{r = 0}^{n - 1} \left (z - \exp \left (i\pi \cdot \frac {1 + 2(2n - 1 - r)}{2n}\right )\right )\right ] \\ & = \left [\prod _{r = 0}^{n - 1} \left (z - \exp \left (i\pi \cdot \frac {1 + 2r}{2n}\right )\right )\right ] \cdot \left [\prod _{r = 0}^{n - 1} \left (z - \exp \left (i\pi \cdot \frac {-1-2r}{2n}\right )\right )\right ] \\ & = \prod _{r = 0}^{n - 1} \left (z - \exp \left (i\pi \cdot \frac {1 + 2r}{2n}\right )\right ) \left (z - \exp \left (i\pi \cdot \frac {-1-2r}{2n}\right )\right ) \\ & = \prod _{r = 0}^{n - 1} \left (z^2 - 2z \cos \left (\frac {2r + 1}{2n} \pi \right ) + 1\right ) \\ & = \prod _{r = 1}^{n} \left (z^2 - 2z \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ). \end {align*}
Let \(z = i\), since \(n\) is even, \(z^{2n} = i^{2n} = (i^2)^n = (-1)^n = 1\). \begin {align*} 2 & = z^{2n} + 1 \\ & = \prod _{r = 1}^{n} \left (i^2 - 2i \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) \\ & = \prod _{r = 1}^{n} 2i \cos \left (\frac {2r - 1}{2n} \pi \right ) \\ & = (2i)^n \prod _{r = 1}^{n} \cos \left (\frac {2r - 1}{2n} \pi \right ) \\ & = 2^n (-1)^{\frac {n}{2}} \prod _{r = 1}^{n} \cos \left (\frac {2r - 1}{2n} \pi \right ), \end {align*}
and therefore \[ \prod _{r = 1}^{n} \cos \left (\frac {2r - 1}{2n} \pi \right ) = 2^{1 - n} (-1)^{-\frac {n}{2}} = 2^{1 - n} (-1)^{\frac {n}{2}}. \]
Notice that in the product where \(n\) is odd, let \(k = \frac {n + 1}{2}\), then the term of this product will be \begin {align*} z^2 - 2z \cos \left (\frac {(2k - 1) \pi }{2n}\right ) + 1 & = z^2 - 2z \cos \left (\frac {(n + 1 - 1)\pi }{2n}\right ) + 1 \\ & = z^2 - 2z \cos \frac {\pi }{2} + 1 \\ & = z^2 + 1. \end {align*}
Therefore, we have \begin {align*} (z^2 + 1) \sum _{r = 0}^{n - 1} (-1)^r z^{2r} & = z^2 + 1 \\ & = \prod _{r = 1}^{n} \left (z^2 - 2z \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) \\ & = \prod _{r = 1}^{\frac {n - 1}{2}} \left (z^2 - 2z \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) (z^2 + 1) \\ & \phantom {=} \prod _{r = \frac {n + 3}{2}}^{n} \left (z^2 - 2z \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) \\ & = \prod _{r = 1}^{\frac {n - 1}{2}} \left (z^2 - 2z \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) (z^2 + 1) \\ & \phantom {=} \prod _{r = 1}^{\frac {n - 1}{2}} \left (z^2 - 2z \cos \left (\frac {2(n + 1 - r) - 1}{2n} \pi \right ) + 1\right ), \end {align*}
and hence \begin {align*} \sum _{r = 0}^{n - 1} (-1)^r z^{2r} & = \prod _{r = 1}^{\frac {n - 1}{2}} \left (z^2 - 2z \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) \left (z^2 - 2z \cos \left (\frac {2(n + 1 - r) - 1}{2n} \pi \right ) + 1\right ) \\ & = \prod _{r = 1}^{\frac {n - 1}{2}} \left (z^2 - 2z \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) \left (z^2 - 2z \cos \left (\frac {2n - 2r + 1}{2n} \pi \right ) + 1\right ) \\ & = \prod _{r = 1}^{\frac {n - 1}{2}} \left (z^2 - 2z \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) \left (z^2 + 2z \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ). \end {align*}
Let \(z = i\), we have \begin {align*} \LHS & = \sum _{r = 0}^{n - 1} (-1)^r i^{2r} \\ & = \sum _{r = 0}^{n - 1} (-1)^r (i^2)^r \\ & = \sum _{r = 0}^{n - 1} (-1)^r (-1)^r \\ & = \sum _{r = 0}^{n - 1} [(-1)(-1)]^r \\ & = \sum _{r = 0}^{n - 1} 1 \\ & = n, \end {align*}
and \begin {align*} \RHS & = \prod _{r = 1}^{\frac {n - 1}{2}} \left (i^2 - 2i \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) \left (i^2 + 2i \cos \left (\frac {2r - 1}{2n} \pi \right ) + 1\right ) \\ & = \prod _{r = 1}^{\frac {n - 1}{2}} (-2i \cos \left (\frac {2r - 1}{2n} \pi \right )) (2i \cos \left (\frac {2r - 1}{2n} \pi \right )) \\ & = \prod _{r = 1}^{\frac {n - 1}{2}} 4 \cos ^2 \left (\frac {2r - 1}{2n} \pi \right ) \\ & = 2^{n - 1} \prod _{r = 1}^{\frac {n - 1}{2}} \cos ^2 \left (\frac {2r - 1}{2n} \pi \right ). \end {align*}
This gives \[ \prod _{r = 1}^{\frac {n - 1}{2}} \cos ^2 \left (\frac {2r - 1}{2n} \pi \right ) = n 2^{1 - n}, \] exactly as desired.