\(% 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}}\)
For \(\omega = \exp \frac {2 \pi i}{n}\), we have for \(k = 0, 1, 2, \ldots , n - 1\), that \(\omega ^k = \exp \frac {2\pi i k}{n}\). Therefore, \[ (\omega ^k)^n = \exp \frac {2\pi i k n}{n} = \exp (2\pi i k) = 1. \]
Also, notice that \(\arg \omega ^k = \frac {2k\pi }{n}\), which means that all \(\omega ^k\)s are different.
This means that \(\omega ^0 = 1, \omega ^1 = 1, \omega ^2, \ldots , \omega ^{n - 1}\) are exactly the \(n\) roots to the polynomial \(z^n - 1\), which has leading coefficient 1.
Therefore, we must have \[ (z - 1)(z - \omega ) \cdots (z - \omega ^{n - 1}) = z^n - 1, \] as desired.
For the following parts, W.L.O.G. let the orientation of the polygon be such that \(X_k = \omega ^k\).
Let \(z\) represent the complex number for \(P\), we have \begin {align*} \prod _{k = 0}^{n - 1} \abs *{P X_k} & = \prod _{k = 0}^{n - 1} \abs *{z - \omega ^k} \\ & = \abs *{\prod _{k = 0}^{n - 1} (z - \omega ^k)} \\ & = \abs *{z^n - 1}. \end {align*}
Since \(P\) is equidistant from \(X_0\) and \(X_1\), we must have that \(P = r \exp \left (\frac {\pi i}{n}\right )\) for some \(r \in \RR \), where \(\abs *{r} = \abs *{OP}\). Therefore, we have \begin {align*} \prod _{k = 0}^{n - 1} \abs *{P X_k} & = \abs *{z^n - 1} \\ & = \abs *{r^n \exp \left (\frac {\pi i}{2}\right ) - 1} \\ & = \abs *{-r^n - 1} \\ & = \abs *{r^n + 1}. \end {align*}
If \(n\) is even, then \(r^n = \abs *{r}^n > 0\), and therefore \(\abs *{r^n + 1} = r^n + 1 = \abs *{r}^n + 1 = \abs *{OP}^n + 1\) as desired.
If \(n\) is odd, and \(r > 0\), then \(r^n = \abs *{r}^n > 0\), and \begin {align*} \LHS & = \abs *{r^n + 1} \\ & = r^n + 1 \\ & = \abs *{r}^n + 1 \\ & = \abs *{OP}^n + 1. \end {align*}
When \(-1 \leq r < 0\), we have \(-1 \leq r^n = -\abs *{r}^n < 0\), and \begin {align*} \LHS & = \abs *{r^n + 1} \\ & = r^n + 1 \\ & = -\abs *{r}^n + 1 \\ & = -\abs *{OP}^n + 1. \end {align*}
When \(r < -1\), we have \(r^n = -\abs *{r}^n < -1\), and \begin {align*} \LHS & = \abs *{r^n + 1} \\ & = -r^n - 1 \\ & = \abs *{r}^n - 1 \\ & = \abs *{OP}^n - 1. \end {align*}
In summary, when \(n\) is odd, we have \[ \prod _{k = 0}^{n - 1} \abs *{P X_k} = \begin {cases} \abs *{OP}^n + 1, & P \text { is in the first quadrant}, \\ -\abs *{OP}^n + 1, & P \text { is in the third quadrant and } \abs *{OP} \leq 1, \\ \abs *{OP}^n - 1, & P \text { is in the third quadrant and } \abs *{OP} > 1. \end {cases} \]
Notice that for a general point \(P\) whose complex number is \(z\), we have \begin {align*} \prod _{k = 1}^{n - 1} \abs *{P X_k} & = (z - \omega ) (z - \omega ^2) \cdots \left (z - \omega ^{n - 1}\right ) \\ & = \frac {z^n - 1}{z - 1} \\ & = 1 + z + z^2 + \cdots + z^{n - 1}. \end {align*}
If we let \(P = X_0\), \(z = 1\), and \(\RHS = n\), just as we desired.