Can a negative power for a binomial expansion be taken as the same expression in the denominator with a positive power?

Question

For instance, will $(x+y)^{-n}$ be equal to $1/((x+y)^n)$? I know this is a silly doubt, but since some $x^{-a} = 1/x^a$, I just wanted to know if this is applicable here also.

Answer 1

Yes, we have $(x+y)^{-n}= \frac{1}{(x+y)^n}$.

Answer 2

As an alternative method, you can take $$(x+y)^{-n}=x^{-n}\bigg(1+\frac{y}{x}\bigg)^{-n}$$ Then your expansion will be: $$\frac{1}{x^n}\bigg[1+(-n)\bigg(\frac{y}{x}\bigg)+\frac{(-n)(-(n+1))}{2!}\bigg(\frac{y}{x}\bigg)^2+\frac{(-n)(-(n+1))(-(n+2))}{3!}\bigg(\frac{y}{x}\bigg)^3+\cdots\bigg]$$