My work:
I wrote it in this form $[(1+w)(1+w^2)(1+w^3)]^k$ where k is the number of repeat of this triplet in a product.
Every three things consecutive factors of $(1+w) ( 1+w^2) ( 1+w^3)$ count as one power to the exponent. I wrote $ k = \frac{q}{3}$ where $q$ is a number of terms in product, so I can group these products which give same value .
Here it is $q=2n$, so
$$ [ (1+w)(1+w^2)(1+w^3)]^{\frac{2n}{3}}$$
But this identity definitely doesn't seem to work for some powers like $n=2$, since
$$ (1+w)(1+w^2)(1+w^3)(1+w^4) = [(1+w)(1+w^2)^{\frac{1}{2}}(1+w^3)^{\frac{1}{2}}]^2$$ which doesn't match with derived identity
We have $$\begin{align} &(1+w)(1+w^2)(1+w^3)\\ =&(1+w)(1+\overline{w})(1+1)\\ =&2(1+w+\overline{w}+w\overline{w})\\ =&2(2+2\Re{w})\\ =&4\left(1+\cos{\frac{2\pi}{3}}\right)\\ =&2 \end{align} $$ Therefore, if $2n=3s+m,\ 0\leq m<3$ we have $$\prod_{k=1}^{2n}(1+w^k)=2^s\prod_{k=1}^{m}(1+w^k)=\begin{cases}2^s,&m\neq1\\2^{s-1}(1+i\sqrt3),&m=1 \end{cases}$$