Is there an explicit formula for power series of $\left(\frac{1+x}{1-x}\right)^n$? I am trying to answer this question.  Per suggestion in one the comments for that question, one might be able use the power series of the terms to arrive at the answer.  However, one of the terms is:
$$\left(\frac{1+x}{1-x}\right)^n$$
I am wondering if there is an explicit power series formula for this expression when $x$ is small.
I cannot see a pattern in the expression for the Taylor series output by Mathematica:
$\left(\frac{1+x}{1-x}\right)^n=1+2 n x+2 n^2 x^2+\frac{2}{3} \left(2 n^3+n\right) x^3+\frac{2}{3} \left(n^4+2 n^2\right) x^4+\frac{2}{15} \left(2 n^5+10 n^3+3 n\right) x^5+\frac{2}{45} \left(2 n^6+20 n^4+23 n^2\right) x^6+\frac{2}{315} \left(4 n^7+70 n^5+196 n^3+45 n\right) x^7+\ldots$

I have likely made a small mistake in this gross derivation, but this is an idea:
$\displaystyle \Big(\frac{1+x}{1-x}\Big)^n=\frac{1}{(1-x)^n}(1+x)^n=(1+x)^n\frac{1}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\frac{1}{1-x}=(x+1)^n\frac{1}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\sum_{i=0}^\infty x^i=\frac{1}{(n-1)!}\sum_{k=0}^n {n\choose k} x^{n-k}\sum_{i=0}^\infty \frac{(i+(n-1)!}{(i)!}x^{i}=\frac{1}{(n-1)!}\sum_{k=0}^\infty {n\choose k}\chi_{k\le n} x^{n-k}\sum_{i=0}^\infty \frac{(i+(n-1)!}{(i)!}x^{i}=\sum_{l=0}^\infty\Bigg(\frac{1}{(n-1)!}\sum_{m=0}^l{n\choose m}\chi_{m\le n}\frac{(l-m-(n-1)!}{(l-m)!}\Bigg)x^l$ 
where the last part uses the Cauchy product. Note that $\chi_{m\le n}=1$ if $m\le n$, $0$ otherwise. You get negative factorials, but  that's why I used the indicator function to make clear that those terms are $0$.
