I am currently trying to express an arbitrary function, $f(x)$ to an arbitrary power, $m$, using a Taylor series $$ \left[f\left(x\right)\right]^m $$ I first considered expressing $f\left(x\right)=\sum_{n=1}^{\infty}{a_n\left(x-b\right)^n}$, where $a_n=\frac{1}{n!}\ \frac{\partial^nf}{\partial x^n}$ and $b$ is the point about which the function is known. I first multiplied the series unto itself m times and got the following:
$$\left[f\left(x\right)\right]^m=\sum_{n=1}^{\infty}{a_{n,1}\left(x-b\right)^n}\sum_{n=1}^{\infty}{a_{n,2}\left(x-b\right)^n}\sum_{n=1}^{\infty}{a_{n,3}\left(x-b\right)^n}\ldots=\ $$
$$=\ldots\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}{a_ia_ja_k\ldots\left(x-b\right)^{i+j+k+\ldots}}\ $$
Where I have $m$ summations upon one another. For example, when $m=3$, the expression looks as follows: $$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}{a_ia_ja_k\left(x-b\right)^{i+j+k}}$$
Is there a simpler way to represent this? I was thinking about using some sort of binomial expansion, but I'm not sure where to start with that. Later I'd like to try to do this a function of two variables, such as $f(x,y)$, but wanna start simple first. Any advice on the latter matter would be greatly appreciated.