Find a power series expansion for the function gf′ on (-1,1) $$ \sum_{k=0}^\infty\ x^k$$ 
has radius of convergence R = 1 
so we may define a function f : (-1,1) by $$ \sum_{k=0}^\infty\ r^k$$ for any r in (-1,1)
Let g(r) = r for all r in (-1,1).
I am asked to find a power series expansion for the function ${gf'}$
on (-1,1). Here ${gf'}$ is the product of
g with the derivative of f .
My question is does power series expansion mean find the Taylor series or I guess Maclaurin series since this in this case it's centered at zero.
When I try to find the Maclaurin series for $x^k$ I get 
$$f'(x)=k(x)^{k-1}$$ $$f''(x)=k(k-1)(x)^{k-2}$$ $$f'''(x)=k(k-1)(k-2)(x)^{k-3}$$
so I think $$f^n(x)=k(k-1)...(kn+1)(x)^{k-n}$$
But I don't know what is meant by ${gf'}$
on (-1,1). Here ${gf'}$ is the product of
g with the derivative of f. 
If anyone could help me understand what the question is asking I would really appreciate it. 
Thank you,
 A: By the fact that we can do termwise differentiation of a power series, if
$$f(x) = \sum_{k=0}^\infty x^k,$$
then
$$f'(x) = \sum_{k=0}^\infty k x^{k-1}.$$
Then we have
$$(g f')(x) = x \sum_{k=0}^\infty kx^{k-1} = \sum_{k=0}^\infty kx^k.$$
Note that one can either prove directly that this has radius of convergence $R = 1$ (for instance using the Cauchy-Hadamard Theorem), or by the property that if two power series are both finite on the interval $(-1,1)$ then so two is their product.
An alternative derivation
This is a simple derivation, we can also check it using the fact that 
$$f(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x}.$$
Therefore 
$$f'(x) = \frac{1}{(1-x)^2},$$ 
and
$$(gf')(x) = \frac{x}{(1-x)^2}.$$
You can now take the Taylor expansion of this and re-derive the above result.
A comment on your solution
A power series is really just a polynomial representation of a function (with technicalities about which functions have valid power series representations).
In your solution you try to compute a Maclaurin / Taylor expansion for the function $h(x) = x^k$. The fact is this is already a power series, so there is no need to do any calculations.
