First, by Sine Exponential Formulation we have: :$\sin \theta = \dfrac 1 {2 i} \paren {e^{i \theta} - e^{-i \theta} }$ Therefore by Power of Product: :$\sin^{2 n} \theta = \dfrac 1 {\paren {2 i}^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$ Now by Power of Product and the Power of Power: :$\dfrac 1 {\paren {2 i}^{2 n} } = \dfrac 1 {2^{2 n} \paren {-1}^n} = \dfrac {\paren {-1}^n} {2^{2 n} }$ Thus: :$\sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$ Now: {{begin-eqn}} {{eqn | l = \paren {e^{i \theta} - e^{-i \theta} }^{2 n}       | r = \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k e^{i \paren {2 n - k} \theta} e^{-i k \theta}       | c = Binomial Theorem }} {{eqn | r = \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k e^{i \paren {2 n - 2 k}\theta}       | c = Power of Product }} {{eqn | r = \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta       | c = Euler's Formula }} {{end-eqn}} So we have: :$\displaystyle \sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {\sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta}$ Now we look at each of the terms in the parentheses: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \map \sin {2 n - 2 k} \theta       | r = \paren {-1}^n \sum_{k \mathop = -n}^n {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta}       | c = replacing $k \mapsto k + n$ }} {{eqn | r = \paren {-1}^n \sum_{k \mathop = -n}^{-1} {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta} + \paren {-1}^n \sum_{k \mathop = 1}^n {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta}       | c = Zeroes of Sine and Cosine }} {{eqn | r = \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \sin 2 k \theta + {2 n \choose n + k} \map \sin {-2 k \theta} }       | c =  }} {{eqn | r = \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n + k} \sin 2 k \theta + {2 n \choose n + k} \map \sin {-2 k \theta} }       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n + k} \sin 2 k \theta - {2 n \choose n + k} \sin 2 k \theta}       | c = Sine Function is Odd }} {{eqn | r = 0 }} {{end-eqn}} We find, for the remaining term: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta       | r = \paren {-1}^n \sum_{k \mathop = -n}^n {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta}       | c = replacing $k \mapsto k + n$ }} {{eqn | r = \paren {-1}^n \sum_{k \mathop = -n}^{-1} {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta} + \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta}       | c = Cosine of Zero is One }} {{eqn | r = \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \cos 2 k \theta + {2 n \choose n + k} \map \cos {-2 k \theta} }       | c =  }} {{eqn | r = \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \cos 2 k \theta + {2 n \choose n + k} \cos 2 k \theta}       | c = Cosine Function is Even }} {{eqn | r = \paren {-1}^n {2 n \choose n} + \paren {-1}^n 2 \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta        | c = Symmetry Rule for Binomial Coefficients }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l = \sin^{2 n} \theta       | r = \frac {\paren {-1}^n} {2^{2 n} } \paren {\paren {-1}^n {2 n \choose n} + \paren {-1}^n 2 \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta} }} {{eqn | r = \frac 1 {2^{2 n} } {2 n \choose n} + \frac 1 {2^{2 n - 1} } \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta  }} {{eqn | r = \frac 1 {2^{2 n} } {2 n \choose n} + \frac {\paren {-1}^n} {2^{2 n - 1} } \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k {2 n \choose k} \map \cos {2 n - 2 k} \theta        | c = replacing $k \mapsto n-k$ }} {{end-eqn}} as required. {{qed}}