So the way I think I should approach this is by getting a result for $n=1,2,3...$ and then examine them. I could easily get the Taylor series expansion for $n=1$, but then I don't really know how to proceed, because I can't always break down a power of a polynomial. And a power of a summation doesn't seem as a legit answer either.
Can anyone help? Thanks
Expanding Daniel Fischer's comment, it is known that for any $x$ such that $|x|<1$: $$\frac{1}{(1-x)^n}=\sum_{k\geq 0}\binom{n+k-1}{n-1}x^k,$$ hence (by replacing $x$ with $-x$): $$\frac{1}{(1+x)^n}=\sum_{k\geq 0}\binom{n+k-1}{n-1}(-1)^k x^k$$ and (by replacing $x$ with $x^2$): $$\frac{1}{(1+x^2)^n}=\sum_{k\geq 0}\binom{n+k-1}{n-1}(-1)^k x^{2k}.$$