<s>[INST] Prove $\left(\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}\right)^n = \cos2n\theta+i\sin2n\theta$ 
Prove
  $$\left(\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}\right)^n = \cos2n\theta+i\sin2n\theta$$

Not sure how to go about this proof, was thinking of putting numerator and denominator in mod arg form straight away and then using binomial expansion but doesn't seem to work out... Then I tried realising the denominator first and ended up with
$$\frac{(1+cis2θ)^{2n} }{ 2^n(1+cos2θ)^n}$$  and not sure where to go from there?? Any help much appreciated, thanks :)
 [/INST] $$\begin{align}
e^{i2\theta n} \cdot (1+e^{-i2\theta})^n
&=e^{i2\theta n}\cdot \left(\binom{n}{0} e^{-i2\theta \cdot 0} + \binom{n}{1} e^{-i2\theta \cdot 1}+\binom{n}{2} e^{-i2\theta \cdot 2} + \dots + \binom{n}{n-1} e^{-i2\theta \cdot (n-1)}+ \binom{n}{n} e^{-i2\theta \cdot n}\right) \\ 
&= \binom{n}{0} e^{i2\theta \cdot(n- 0)} + \binom{n}{1} e^{i2\theta \cdot (n- 1)}+\binom{n}{2} e^{i2\theta \cdot (n- 2)} + \dots + \binom{n}{n-1} e^{i2\theta \cdot (n-(n-1))}+ \binom{n}{n} e^{i2\theta \cdot(n- n)} \\ 
&= \binom{n}{0} e^{i2\theta \cdot n} + \binom{n}{1} e^{i2\theta \cdot (n- 1)}+\binom{n}{2} e^{i2\theta \cdot (n- 2)} + \dots + \binom{n}{n-1} e^{i2\theta \cdot 1}+ \binom{n}{n} e^{i2\theta \cdot 0}\\
&=\binom{n}{n} e^{i2\theta \cdot n} + \binom{n}{n-1} e^{i2\theta \cdot (n- 1)}+\binom{n}{n-2} e^{i2\theta \cdot (n- 2)} + \dots + \binom{n}{1} e^{i2\theta \cdot 1}+ \binom{n}{0} e^{i2\theta \cdot 0}\\
&=(1+e^{i2\theta})^n
\end{align}$$
Thus, 
$$
e^{i2\theta n}= \frac{
(1+e^{i2\theta})^n}{(1+e^{-i2\theta})^n} =  \left(\frac{
1+e^{i2\theta}}{1+e^{-i2\theta}} \right)^n=\left(\frac{
1+\cos2\theta +i \sin2\theta}{1+\cos2\theta -i \sin2\theta} \right)^n \ .$$
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