Let $e_j = \cos(2 \pi j x) + i \sin( 2 \pi j x)$ (the $j^{th}$ fourier basis term on the unit circle $[0,1])$
How can I show $2^{2n} \cos ^ {2n} (\pi x) = \sum _{k=0} ^{2n} {2n \choose k } e_{n - k}$
Attempt: The rhs can be rewritten $\sum _{k = -n} ^n {2n \choose n+k}e_{k}$ so that it is equivalent to compute the fourier coefficients of $2^{2n} \cos ^ {2n} (\pi x)$.
The $k^{th}$ coefficient is given $\int 2^{2n} \cos ^ {2n} (\pi x) (\cos(2 \pi k x) - i \sin(2 \pi k x)) \,dx = \int 2^{2n} \cos ^ {2n} (\pi x) \cos(2 \pi k x) \,dx$ with the $\sin$ terms dropping out because of symmetry about $0$.
$2^{2n} \cos^{2n}(\pi x) = 2^n(1+\cos(2 \pi x))^n = 2^n \sum_{k= 0}^n {n \choose k} \cos ^n (2 \pi x)$
At this point I am getting scared of the powers of cosine so I think I'm going down the wrong path.
This is just the plain old binomial theorem $$ 2^{2n}\cos^{2n}(\pi x)=\left(e^{i\pi x}+e^{-i\pi x}\right)^{2n} = \sum_{k=0}^{2n}\binom{2n}{k}\cdot e^{i\pi x·(2n-k)}·e^{-i\pi x\cdot k} $$ from which the statement follows.