Unifying the product of odd powers of function with the product of even powers of the same function in one product. Assume that
\begin{equation}
\begin{split}
f_n(x)&= \begin{cases} \big(g(x)\big)\cdot \big(g(x)\big)^3\cdot \big(g(x)\big)^5\cdots \big(g(x)\big)^{n-1},& n \ \text{even},\\
\big(g(x)\big)^2\cdot \big(g(x)\big)^4\cdot \big(g(x)\big)^6\cdots \big(g(x)\big)^{n-1},& n \ \text{odd}, \end{cases}\\
&=\begin{cases}\displaystyle\prod^{n-1}_{k\ \ odd} \big(g(x)\big)^{k}  ,& n \ \text{even},\\
\displaystyle\prod^{n-1}_{k\ \ even}\big(g(x)\big)^{k},& n \ \text{odd}, \end{cases}
\end{split}
\end{equation}
where $g(x)$ is any polynomial of order $m$(with real coefficient). I beleive that it's possible to unify the two cases($n$ odd and even) in one product like $\prod^{n-1}_{k=1}(\cdots)$ or $\prod^{n-2}_{k=0}(\cdots)$ $\big($or like $\prod^{n/2}_{k=1}(\cdots)$ or $\prod^{(n/2)-1}_{k=0}(\cdots)$$\big)$ which gives the desired $f_n(x)$ for each $n$. I hope someone helps me to figure it out.

you can do much better actually. Let me start with $n=2N+1$ odd, then
$$
f_n(x)=g(x)^{\sum_{k=0}^{N}(2k)}=g(x)^{2\sum_{k=0}^{N-1}k}=g(x)^{N(N+1)}
$$
and for $n=2N$ even 
$$
f_n(x)=g(x)^{\sum_{k=0}^{N-1}(2k+1)}=g(x)^{2\sum_{k=0}^{N-1}k+N}=g(x)^{N(N+2)}
$$
There are many ways to rewrite this in a single line, one of them is
$$
f_n=g(x)^{\lfloor n/2\rfloor\left(\lfloor n/2\rfloor+1+\frac{1+(-1)^n}{2}\right)}
$$
