I was trying to factor $a+b$: $$a+b=\left(a^{1/2}\right)^2+\left(b^{1/2}\right)^2$$ $$=\left(a^{1/2}\right)^2-\left(ib^{1/2}\right)^2$$ $$=\left(a^{1/2}+ib^{1/2}\right)\left(a^{1/2}+\left(-ib^{1/2}\right)\right)$$ So by recursion:
$$a+b=\prod_{m=1}^{2^n}\left(a^{1/2^n}+(-1)^{(m-1)/2^{n-1}}i^{1/2^{n-1}} b^{1/2^n}\right)$$ $$=\prod_{m=1}^{2^n}\left(a^{\frac{1}{2^n}}+e^{i\pi\left({(2m-1)/2^n}\right)} b^{1/2^n}\right)$$
Is this valid? If so, is: $$a+b=\lim_{n\to\infty}\left(\prod_{m=1}^{2^n}\left(a^{{1/2^n}}+e^{i\pi\left({(2m-1)/2^n}\right)} b^{1/2^n}\right)\right)$$ also valid?
Note: I have tested this in Maple 2018 and it seems to hold true: for $a=5,b=10,n=15$ I get $14.999658+0.000062 i$ using evalf.
EDIT #1:
By pairing factors with complex conjugates, I was able to find an expression without the use of complex numbers:
$$a^{2^n}+b^{2^n}=\prod_{m=1}^{2^{n-1}}\left(a^2+b^2+2\cos\left(\frac{2m-1}{2^n}\pi\right)ab\right)$$
Hence, by letting $b=ka$:
$$k^{2n}+1=\prod_{m=1}^{n}\left(k^2+2k\cos\left(\frac{2m-1}{2n}\pi\right)+1\right)$$
EDIT #2:
By letting $a=b$, I derived the following result:
$$\prod_{m=1}^{2^{n-1}}\left(1+\cos\left(\frac{2m-1}{2^n}\pi\right)\right)=2^{1-2^{n-1}}$$
which also seems to hold true empirically.
EDIT #3:
Another identity:
$$\prod_{m=1}^{2^{n-1}}\cos^2\left(\frac{2m-1}{2^{n+1}}\pi\right)=\frac{1}{2^{2^n-1}}$$
Generalising this gives:
$$\prod_{m=1}^{n}\cos\left(\frac{2m-1}{4n}\pi\right)=\frac{\sqrt{2}}{2^n}$$