Proving $ z^n + \frac{1}{z^n} = 2\cos(n\theta) $ for $z = \cos (\theta) + i\sin(\theta)$ 

Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n +  {1\over z^n} = 2\cos(n\theta) $$



What I have attempted
If $$  z = \cos (\theta) + i\sin(\theta) $$
then $$ z^n = \cos (n\theta) + i\sin(n\theta) $$
$$ z^n +  {1\over z^n} $$
$$ \cos (n\theta) + i\sin(n\theta) +  {1\over \cos (n\theta) + i\sin(n\theta)} $$
$$ (\cos (n\theta) + i\sin(n\theta))\cdot(\cos (n\theta) + i\sin(n\theta)) + 1 $$
$$ \left[ {(\cos (n\theta) + i\sin(n\theta))\cdot(\cos (n\theta) + i\sin(n\theta)) + 1\over \cos (n\theta) + i\sin(n\theta)}  \right]  $$
$$ \left[ {\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + 1\over \cos (n\theta) + i\sin(n\theta)}  \right]  $$
Now this is where I am stuck.. I tried to use a double angle identity but I can't eliminate the imaginary part..

setting the value of $z=\cos\theta+i\sin\theta$, $$z^n+\frac{1}{z^n}$$$$=z^n+z^{-n}$$
$$=(\cos\theta+i\sin\theta)^n+(\cos\theta+i\sin\theta)^{-n}$$
Using d-Moivre's theorem, 
$$=\cos(n\theta)+i\sin(n\theta)+\cos(-n\theta)+i\sin(-n\theta)$$
$$=\cos(n\theta)+i\sin(n\theta)+\cos(n\theta)-i\sin(n\theta)$$
$$=2\cos(n\theta)$$
